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First, find the time of flight \(t\) using the vertical motion.
Using \(s = ut + \frac{1}{2}at^2\) vertically:
\(h = 0 + \frac{1}{2}gt^2\)
\(t = \sqrt{\frac{2h}{g}}\)

Next, use the horizontal motion to find the distance \(d\):
\(d = v \times t = v \sqrt{\frac{2h}{g}}\)

1 mark for the correct option C.
[1] Correctly identifies that vertical motion gives \(t = \sqrt{\frac{2h}{g}}\) and horizontal distance is \(d = vt\).

Calculate the useful power output:
\(P_{\text{out}} = Fv = mgv = 120\text{ kg} \times 9.81\text{ m s}^{-2} \times 0.50\text{ m s}^{-1} = 588.6\text{ W}\)

Calculate the power input using efficiency:
\(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}}\)
\(0.60 = \frac{588.6\text{ W}}{P_{\text{in}}}\)
\(P_{\text{in}} = \frac{588.6\text{ W}}{0.60} = 981\text{ W}\)

This rounds to \(980\text{ W}\).

1 mark for the correct option C.
[1] Correct calculation of useful power output as \(588.6\text{ W}\) and division by \(0.60\) to obtain \(980\text{ W}\).

The Young Modulus is given by:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta x / L} = \frac{FL}{A\Delta x}\)
Rearranging for force \(F\):
\(F = \frac{E A \Delta x}{L} = \frac{E \pi d^2 \Delta x}{4 L}\)
So, \(F \propto \frac{d^2}{L}\).
For the second wire, \(d_2 = 2d\) and \(L_2 = 2L\):
\(F_2 \propto \frac{(2d)^2}{2L} = \frac{4d^2}{2L} = 2 \left(\frac{d^2}{L}\right)\)
Therefore, the required force is \(2F\).

1 mark for the correct option C.
[1] Correctly relates force to wire geometry as \(F \propto \frac{d^2}{L}\) and determines that doubling both results in twice the force.

Let the acceleration of the system be \(a\) and tension in the string be \(T\).
Using Newton's Second Law for each block:
For the \(3.0\text{ kg}\) block (moving downwards): \(3.0g - T = 3.0a\)
For the \(1.0\text{ kg}\) block (moving upwards): \(T - 1.0g = 1.0a\)

Adding these two equations:
\(2.0g = 4.0a \implies a = 0.50g = 4.905\text{ m s}^{-2}\)

Substitute \(a\) back into the equation for the \(1.0\text{ kg}\) block:
\(T - 1.0(9.81) = 1.0(4.905)\)
\(T = 9.81 + 4.905 = 14.715\text{ N} \approx 15\text{ N}\)

1 mark for the correct option B.
[1] Correctly sets up equation of motion to find acceleration \(a = 4.9\text{ m s}^{-2}\) and calculates tension as \(15\text{ N}\).

At terminal velocity, the upward forces balance the downward force:
\(\text{Drag Force} + \text{Upthrust} = \text{Weight}\)
\(6\pi\eta r v + \frac{4}{3}\pi r^3 \rho_{\text{fluid}} g = \frac{4}{3}\pi r^3 \rho_{\text{metal}} g\)
\(6\pi\eta r v = \frac{4}{3}\pi r^3 g (\rho_{\text{metal}} - \rho_{\text{fluid}})\)
Therefore, \(v \propto r^2\).

When the radius is doubled to \(2r\):
\(v_2 \propto (2r)^2 = 4r^2\)
So, the new terminal velocity is \(4v\).

1 mark for the correct option D.
[1] Correctly identifies that terminal velocity \(v\) is proportional to the square of the radius \(r^2\) from Stokes' Law.

During upward motion, both gravity and air resistance act downwards on the ball. Therefore, the net downward force is \(mg + F_D\), where \(F_D\) is the drag force.
This gives a deceleration magnitude of \(g + \frac{F_D}{m}\).
Because the speed is highest immediately after release, the drag force \(F_D\) is at its maximum at this point, resulting in the maximum acceleration (deceleration) magnitude. At the top, speed is zero so \(F_D = 0\) and acceleration is \(g\). On the way down, the net force is \(mg - F_D\), meaning acceleration is less than \(g\).

1 mark for the correct option A.
[1] Correctly applies force analysis showing that drag and weight act in the same direction on the way up, making total force maximum when speed (and hence drag) is greatest.

The area under the loading curve (\(W_1\)) represents the work done on the rubber band to stretch it.
The area under the unloading curve (\(W_2\)) represents the work done by the rubber band as it returns to its original length.
The difference between these two areas, \(W_1 - W_2\), is the energy lost as heat (thermal energy) in the rubber band due to internal friction.

1 mark for the correct option B.
[1] Correctly identifies that the work done during loading is \(W_1\), work recovered during unloading is \(W_2\), and the hysteresis loss is the difference.

Taking the initial direction of travel (to the right) as positive:
Initial velocity \(u = +3.0\text{ m s}^{-1}\)
Final velocity \(v = -2.0\text{ m s}^{-1}\)

Change in momentum \(\Delta p = m(v - u) = 0.20\text{ kg} \times (-2.0 - 3.0)\text{ m s}^{-1} = -1.0\text{ kg m s}^{-1}\)

Using Newton's Second Law:
\(F = \frac{\Delta p}{\Delta t} = \frac{-1.0\text{ kg m s}^{-1}}{0.10\text{ s}} = -10\text{ N}\)

The magnitude of the force is \(10\text{ N}\).

1 mark for the correct option C.
[1] Calculates change in momentum correctly as \(1.0\text{ kg m s}^{-1}\) taking direction into account, and divides by \(0.10\text{ s}\) to get \(10\text{ N}\).

Since the block is sliding down the slope at a constant speed, the acceleration is zero and the net force parallel to the slope is zero.

The component of the weight acting down the slope is:
\(F_{\parallel} = mg \sin\theta\)

Therefore, the frictional force \(F_f\) acting up the slope must balance this component:
\(F_f = mg \sin\theta\)

The rate of doing work against friction is the power dissipated by the frictional force:
\(\text{Power} = \text{Force} \times \text{velocity} = F_f \times v = mgv \sin\theta\)

1 mark for the correct option (B).

- Award 1 mark for identifying that the frictional force is equal to the component of weight down the slope, \(mg \sin\theta\), and multiplying by speed \(v\) to obtain \(mgv \sin\theta\).

For a negative temperature coefficient (NTC) thermistor, resistance is inversely related to temperature. Therefore, when the temperature decreases, the resistance of the thermistor increases.

In a series potential divider circuit, the total e.m.f. is shared between the components in proportion to their resistance. As the resistance of the thermistor increases, it takes a larger share of the potential difference. Consequently, the potential difference across the fixed resistor decreases, which causes the voltmeter reading to decrease.

1 mark for the correct option (C).

- Award 1 mark for recognizing that a decrease in temperature increases the resistance of an NTC thermistor, which leads to a decrease in the voltmeter reading across the fixed resistor.

First, calculate the volume of water hitting the wall per second: \(\frac{\Delta V}{\Delta t} = A v = 2.5 \times 10^{-4}\text{ m}^2 \times 8.0\text{ m s}^{-1} = 2.0 \times 10^{-3}\text{ m}^3\text{ s}^{-1}\). Next, calculate the mass of water hitting the wall per second: \(\frac{\Delta m}{\Delta t} = \rho \frac{\Delta V}{\Delta t} = 1000\text{ kg m}^{-3} \times 2.0 \times 10^{-3}\text{ m}^3\text{ s}^{-1} = 2.0\text{ kg s}^{-1}\). Finally, calculate the force using Newton's second law where the force is the rate of change of momentum: \(F = \frac{\Delta p}{\Delta t} = \frac{\Delta m}{\Delta t} \times \Delta v = 2.0\text{ kg s}^{-1} \times 8.0\text{ m s}^{-1} = 16\text{ N}\).

MP1: Use of \(\rho A v\) to find the mass flow rate of water (e.g., \(2.0\text{ kg s}^{-1}\)). (1 mark) MP2: Use of \(F = v \frac{\Delta m}{\Delta t}\) (or equivalent momentum change calculation). (1 mark) MP3: Correct calculation of force to give \(16\text{ N}\). (1 mark)

First, calculate the extension \(\Delta x\) of the wire using the Young Modulus formula \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F L}{A \Delta x}\). Rearranging for extension gives: \(\Delta x = \frac{F L}{A E} = \frac{90\text{ N} \times 2.2\text{ m}}{4.5 \times 10^{-7}\text{ m}^2 \times 2.0 \times 10^{11}\text{ Pa}} = 2.2 \times 10^{-3}\text{ m}\). Then, calculate the elastic strain energy stored using \(E_{\text{el}} = \frac{1}{2} F \Delta x = 0.5 \times 90\text{ N} \times 2.2 \times 10^{-3}\text{ m} = 0.099\text{ J}\) (or \(0.10\text{ J}\)).

MP1: Use of \(E = \frac{F L}{A \Delta x}\) to express extension. (1 mark) MP2: Correct calculation of the extension of the wire as \(2.2 \times 10^{-3}\text{ m}\). (1 mark) MP3: Use of \(E_{\text{el}} = \frac{1}{2} F \Delta x\) to yield \(0.099\text{ J}\) or \(0.10\text{ J}\). (1 mark)

(a) Vertical component of velocity \(u_y = u \sin(\theta) = 8.5 \sin(35^\circ) = 4.88\text{ m s}^{-1}\), which rounds to \(4.9\text{ m s}^{-1}\). (b) Using \(s = u_y t + \frac{1}{2} a_y t^2\), and taking upwards as positive: \(-1.2 = 4.88 t - 4.905 t^2\). Rearranging gives the quadratic equation \(4.905 t^2 - 4.88 t - 1.2 = 0\). Solving for \(t\) using the quadratic formula gives \(t = \frac{4.88 \pm \sqrt{(-4.88)^2 - 4(4.905)(-1.2)}}{2(4.905)} = 1.20\text{ s}\) (discarding the negative root). (c) By conservation of energy, the final kinetic energy \(E_{k,\text{final}}\) is the sum of initial kinetic energy and initial gravitational potential energy: \(E_{k,\text{final}} = \frac{1}{2} m u^2 + m g h = \frac{1}{2}(0.045)(8.5)^2 + (0.045)(9.81)(1.2) = 1.626\text{ J} + 0.530\text{ J} = 2.16\text{ J}\).

(a) [1M] for using \(u_y = u \sin(\theta)\); [1M] for obtaining \(4.88\text{ m s}^{-1}\) and concluding it is approximately \(4.9\text{ m s}^{-1}\). (b) [1M] for selecting \(s = u t + \frac{1}{2} a t^2\) with correct signs; [1M] for substituting correct values into the equation; [1M] for rearranging into standard quadratic form; [1M] for final answer of \(1.20\text{ s}\) (accept \(1.2\text{ s}\)). (c) [1M] for using conservation of energy equation or calculating final components of velocity; [1M] for final kinetic energy of \(2.16\text{ J}\) (accept \(2.2\text{ J}\)).

(a) \(A = \frac{\pi d^2}{4} = \frac{\pi (0.80 \times 10^{-3})^2}{4} = 5.03 \times 10^{-7}\text{ m}^2\). (b) The tensile force \(F = m g = 6.0 \times 9.81 = 58.86\text{ N}\). Young modulus \(E = \frac{F L}{A \Delta L}\), so \(\Delta L = \frac{F L}{A E} = \frac{58.86 \times 2.5}{(5.03 \times 10^{-7}) \times (2.0 \times 10^{11})} = 1.46 \times 10^{-3}\text{ m}\) (or \(1.46\text{ mm}\)). (c) Since Young modulus \(E \propto \frac{1}{\Delta L}\) for constant dimensions and force, and the Young modulus of brass is half that of steel, the extension of the brass wire will be doubled, which is \(2.9\text{ mm}\).

(a) [1M] for using \(A = \pi r^2\) or \(A = \frac{\pi d^2}{4}\); [1M] for correct calculation of area \(5.0 \times 10^{-7}\text{ m}^2\). (b) [1M] for calculating weight \(F = 59\text{ N}\); [1M] for recalling \(E = \frac{F L}{A \Delta L}\); [1M] for algebraic rearrangement for \(\Delta L\); [1M] for correct answer of \(1.46 \times 10^{-3}\text{ m}\) (accept \(1.5 \times 10^{-3}\text{ m}\)). (c) [1M] for identifying that Young modulus is inversely proportional to extension; [1M] for stating the extension doubles to \(2.9\text{ mm}\) (or \(3.0\text{ mm}\)).

(a) By conservation of linear momentum: \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\). Taking the initial direction of the car as positive: \((0.15 \times 2.2) + 0 = (0.15 \times -0.50) + (0.25 \times v_2)\). \(0.33 = -0.075 + 0.25 v_2 \implies 0.405 = 0.25 v_2 \implies v_2 = 1.62\text{ m s}^{-1}\). (b) Initial Kinetic Energy: \(E_{k,i} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} (0.15)(2.2)^2 = 0.363\text{ J}\). Final Kinetic Energy: \(E_{k,f} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} (0.15)(-0.50)^2 + \frac{1}{2} (0.25)(1.62)^2 = 0.01875 + 0.32805 = 0.3468\text{ J}\). Since \(E_{k,f} < E_{k,i}\), kinetic energy is not conserved, hence the collision is inelastic. (c) Average force on the toy car: \(F = \frac{\Delta p}{\Delta t} = \frac{m_1(v_1 - u_1)}{\Delta t} = \frac{0.15(-0.50 - 2.2)}{0.080} = -5.06\text{ N}\). The magnitude of the average force is \(5.1\text{ N}\).

(a) [1M] for applying conservation of momentum; [1M] for correct substitution with appropriate signs; [1M] for \(1.62\text{ m s}^{-1}\) (or \(1.6\text{ m s}^{-1}\)). (b) [1M] for calculating initial kinetic energy; [1M] for calculating total final kinetic energy; [1M] for comparing values and concluding that collision is inelastic because kinetic energy is not conserved. (c) [1M] for using \(F = \frac{\Delta p}{\Delta t}\) with correct sign change in momentum; [1M] for correct calculation of force magnitude as \(5.1\text{ N}\) (accept negative sign).

(a) Take moments about point A: The weight of the uniform board acts at its midpoint, \(2.0\text{ m}\) from A: \(W_b = 35 \times 9.81 = 343.35\text{ N}\). The weight of the diver acts at \(4.0\text{ m}\) from A: \(W_d = 70 \times 9.81 = 686.7\text{ N}\). Clockwise moments about A: \((W_b \times 2.0) + (W_d \times 4.0) = (343.35 \times 2.0) + (686.7 \times 4.0) = 686.7 + 2746.8 = 3433.5\text{ N m}\). Counter-clockwise moments about A: \(R_B \times 1.2\). Since in equilibrium: \(R_B \times 1.2 = 3433.5 \implies R_B = 2861.25\text{ N} \approx 2860\text{ N}\). (b) For vertical equilibrium, total upward forces equal total downward forces: \(R_B + R_A = W_b + W_d\). Assuming \(R_A\) acts upwards: \(2861.25 + R_A = 343.35 + 686.7 = 1030.05\text{ N}\). This gives \(R_A = 1030.05 - 2861.25 = -1831.2\text{ N}\). The negative sign shows that \(R_A\) acts downwards with a magnitude of \(1830\text{ N}\). (c) The clockwise moments about support B from both the diver and the board's own weight exceed any self-balancing upward force. Support A must pull downwards to provide a counter-clockwise moment about pivot B to maintain rotational equilibrium.

(a) [1M] for identifying correct distances for weights (\(2.0\text{ m}\) and \(4.0\text{ m}\)); [1M] for applying the principle of moments about A; [1M] for calculating \(R_B = 2860\text{ N}\) (accept \(2.9\text{ kN}\)). (b) [1M] for applying vertical force equilibrium; [1M] for calculating magnitude of \(1830\text{ N}\) (accept \(1.8\text{ kN}\)); [1M] for correctly identifying direction as downwards. (c) [1M] for noting that gravity creates a clockwise moment about pivot B; [1M] for concluding that support A must exert a downward force to create a counter-balancing counter-clockwise moment about B.

(a) The three forces acting on the ball are: Weight (acting vertically downwards), Upthrust (acting vertically upwards), and Viscous drag (acting vertically upwards). (b) Volume of the spherical ball: \(V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.5 \times 10^{-3})^3 = 1.414 \times 10^{-8}\text{ m}^3\). Upthrust equal to the weight of fluid displaced: \(U = V \rho_{\text{fluid}} g = (1.414 \times 10^{-8}) \times 1260 \times 9.81 = 1.747 \times 10^{-4}\text{ N} \approx 1.75 \times 10^{-4}\text{ N}\). (c) Weight of the ball: \(W = V \rho_{\text{steel}} g = (1.414 \times 10^{-8}) \times 7800 \times 9.81 = 1.082 \times 10^{-3}\text{ N}\). At terminal velocity, \(W = U + F_D\), so viscous drag \(F_D = W - U = 1.082 \times 10^{-3} - 1.747 \times 10^{-4} = 9.073 \times 10^{-4}\text{ N}\). By Stokes' Law: \(F_D = 6 \pi \eta r v \implies v = \frac{F_D}{6 \pi \eta r} = \frac{9.073 \times 10^{-4}}{6 \pi \times 1.49 \times (1.5 \times 10^{-3})} = 0.0215\text{ m s}^{-1}\).

(a) [1M] for weight downwards; [1M] for both upthrust and viscous drag upwards. (b) [1M] for calculating sphere volume \(1.41 \times 10^{-8}\text{ m}^3\); [1M] for formula \(U = V \rho g\); [1M] for correct final answer \(1.75 \times 10^{-4}\text{ N}\). (c) [1M] for calculating the ball's weight \(1.08 \times 10^{-3}\text{ N}\); [1M] for using \(F_D = W - U\) and Stokes' Law; [1M] for correct terminal velocity of \(0.0215\text{ m s}^{-1}\) (accept \(0.022\text{ m s}^{-1}\)).

(a) Due to air resistance and friction between the block and the track, some of the initial gravitational potential energy (GPE) of the block is transferred to thermal energy rather than kinetic energy (KE). Consequently, the block exits the track with a lower horizontal velocity, resulting in a smaller horizontal distance \(d\). To determine this thermal energy, we can measure the vertical launch height \(H\) and the horizontal distance \(d\). The time of flight \(t\) is calculated using \(H = \frac{1}{2}gt^2\). The exit velocity is \(v = d/t\). The thermal energy loss is the difference between the initial GPE at the top of the track (\(mgh\)) and the exit KE (\(\frac{1}{2}mv^2\)).

(b) Step 1: Calculate flight time \(t\) using vertical motion:
\(H = \frac{1}{2} g t^2 \implies 0.85 = 0.5 \times 9.81 \times t^2\)
\(t^2 = 0.1733 \implies t = 0.416 \text{ s}\)

Step 2: Calculate horizontal exit velocity \(v\):
\(v = \frac{d}{t} = \frac{0.52}{0.416} = 1.25 \text{ m s}^{-1}\)

Step 3: Calculate initial GPE and actual exit KE:
\(E_p = mgh = 0.15 \times 9.81 \times 0.45 = 0.662 \text{ J}\)
\(E_k = \frac{1}{2} m v^2 = 0.5 \times 0.15 \times (1.25)^2 = 0.117 \text{ J}\)

Step 4: Calculate energy transferred to thermal energy:
\(\Delta E = E_p - E_k = 0.662 - 0.117 = 0.545 \text{ J} \approx 0.55 \text{ J}\)

(c) Modifications:
1. Use a carbon paper or sand tray on the floor where the block lands to mark the exact landing position, reducing human reaction time/visual error in locating the landing spot.
2. Increase the launch height \(H\) to increase the time of flight and therefore the horizontal distance \(d\), which reduces the percentage uncertainty in measuring \(d\) with a metre rule.
3. Use a video camera with a scale/grid in the background to capture the landing frame-by-frame, improving the accuracy of the landing position measurement.

Part (a) [Max 3 marks]:
- Explains that friction/air resistance converts GPE to thermal energy (1)
- Identifies that this leads to a smaller horizontal exit velocity (1)
- Explains how finding the flight time from vertical drop allows exit velocity and thus final KE to be determined (1)

Part (b) [Max 4 marks]:
- Calculation of flight time: \(t = 0.416 \text{ s}\) (1)
- Calculation of horizontal velocity: \(v = 1.25 \text{ m s}^{-1}\) (1)
- Calculation of initial GPE (\(0.66 \text{ J}\)) and actual KE (\(0.12 \text{ J}\)) (1)
- Correct calculation of energy loss: \(0.54 \text{ J}\) to \(0.55 \text{ J}\) (1)

Part (c) [Max 3 marks]:
- Identifies a valid method to record the landing spot precisely (e.g., carbon paper, video analysis, sand) (1)
- Explains how this decreases absolute uncertainty (1)
- Suggests repeating and averaging to find a mean, or increasing \(d\) to reduce percentage uncertainty (1)

(a) 1. Measure the original length \(L\) using a metre rule.
2. Measure the diameter \(d\) of the wire using a micrometer screw gauge. To minimise uncertainty in cross-sectional area, take measurements at several different positions and orientations along the wire, then calculate a mean diameter.
3. Apply known masses (forces \(F\)) to the end of the wire and measure the corresponding extensions \(\Delta x\) using a vernier scale or a marker on the wire against a metre rule.
4. Plot a graph of force vs extension (or stress vs strain) to determine the Young modulus from the gradient.

(b) Step 1: Calculate the cross-sectional area \(A\):
\(A = \frac{\pi d^2}{4} = \frac{\pi (0.46 \times 10^{-3})^2}{4} = 1.662 \times 10^{-7} \text{ m}^2\)

Step 2: Use the Young modulus formula:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F L}{A \Delta x}\)
\(\Delta x = \frac{F L}{A E} = \frac{35.0 \times 2.20}{1.662 \times 10^{-7} \times 1.2 \times 10^{11}} = \frac{77.0}{19944} = 3.86 \times 10^{-3} \text{ m} = 3.9 \text{ mm}\)

(c) The stress-strain graph starts with a straight line through the origin up to the limit of proportionality, followed closely by the elastic limit. Beyond this, it curves to the yield point where large plastic deformation occurs with minimal additional stress. At the yield point, the internal crystal planes of the copper slide past one another due to the movement of dislocations, leading to permanent plastic deformation.

Part (a) [Max 4 marks]:
- Identifies micrometer used for diameter and metre rule for original length (1)
- States that diameter is measured at multiple positions/orientations and averaged to reduce area uncertainty (1)
- Describes applying forces and measuring extension (e.g., using a marker/scale) (1)
- Identifies that the gradient of a suitable graph (stress-strain or F-extension) is used to find the Young modulus (1)

Part (b) [Max 3 marks]:
- Correct calculation of area \(A = 1.66 \times 10^{-7} \text{ m}^2\) (1)
- Correct rearrangement of Young modulus formula for extension \(\Delta x\) (1)
- Calculation of extension as \(3.9 \times 10^{-3} \text{ m}\) or \(3.9 \text{ mm}\) (allow 3.86 mm) (1)

Part (c) [Max 3 marks]:
- Sketch shows linear region followed by a curve with a yield point (1)
- Correctly labels limit of proportionality, elastic limit, and yield point (1)
- Explains that at the yield point, crystal planes slide past each other / dislocations move, causing permanent/plastic deformation (1)

The drift velocity is given by the formula \(I = n A v e\), which means \(v = I / (n A e)\). Since the second wire has twice the diameter, its cross-sectional area is four times larger: \(A_2 = 4A\). Both wires are made of the same material, so the charge carrier density \(n\) is the same. The drift velocity in the second wire is \(v_2 = 3I / (n \times 4A \times e) = 0.75 \times (I / (n A e)) = 0.75 v\).

1 mark for correct option B. Correctly relates the area to the diameter ratio and applies the drift velocity formula.

Using the diffraction grating equation \(d \sin\theta = n \lambda\), for the third-order maximum we have \(d \sin\theta = 3\lambda\), so \(\lambda = \frac{d \sin\theta}{3}\). For the fourth-order maximum with the new wavelength, the equation is \(d \sin\phi = 4 \left(\frac{2}{3}\lambda\right) = \frac{8}{3}\lambda\). Substituting \(\lambda\) into this gives \(d \sin\phi = \frac{8}{3} \left(\frac{d \sin\theta}{3}\right)\), which simplifies to \(\sin\phi = \frac{8}{9}\sin\theta\). Therefore, the angle is \(\phi = \sin^{-1}\left(\frac{8}{9}\sin\theta\right)\).

1 mark for correct option B. Employs the grating equation for both cases and eliminates the wavelength and slit spacing.

The total resistance in the circuit is \(R_{total} = r + 3r = 4r\). The current in the circuit is \(I = \frac{E}{4r\)}. The power dissipated in the external resistor is \(P_{ext} = I^2 \times 3r\). The total electrical power generated by the cell is \(P_{total} = E I = I^2 \times 4r\). The ratio of the external power to the total power is therefore \(\frac{P_{ext}}{P_{total}} = \frac{I^2 \times 3r}{I^2 \times 4r} = \frac{3}{4} = 0.75\).

1 mark for correct option D. Utilizes the current and power formulas to determine the power ratio.

According to Einstein's photoelectric equation, \(hf = \phi + E_k\) where \(\phi\) is the work function of the metal. For frequency \(2f\), the new maximum kinetic energy \(E_k'\) is given by \(E_k' = 2hf - \phi\). Substituting \(hf = \phi + E_k\) gives \(E_k' = 2(\phi + E_k) - \phi = 2E_k + \phi\). Since the work function \(\phi\) is always a positive quantity for photoelectron emission to occur, \(E_k'\) must be greater than \(2E_k\).

1 mark for correct option C. Applies the photoelectric equation to demonstrate that the new maximum kinetic energy is \(2E_k + \phi\).

The resistance of a wire is given by \(R = \rho \frac{\text{length}}{\text{area}}\). The area of wire X is \(A_X = \pi r^2\), so its resistance is \(R_X = \rho \frac{L}{\pi r^2}\). The area of wire Y is \(A_Y = \pi (2r)^2 = 4\pi r^2\), so its resistance is \(R_Y = \rho \frac{2L}{4\pi r^2} = \frac{1}{2} R_X\). Since the wires are connected in series, the current \(I\) through them is the same. The potential difference across each wire is given by \(V = I R\). Therefore, the ratio of the potential differences is \(V_X / V_Y = R_X / R_Y = 2\).

1 mark for correct option D. Calculates the resistance ratio using resistivity and dimensions, then applies Ohm's law in series.

The critical angle \(c\) at the boundary between two media is given by \(\sin c = \frac{n_2}{n_1}\), where \(n_1\) is the refractive index of the denser medium (glass, \(1.52\)) and \(n_2\) is the refractive index of the rarer medium. Rearranging this gives \(n_2 = n_1 \sin c\). Substituting the given values: \(n_2 = 1.52 \times \sin(58.0^\circ) \approx 1.52 \times 0.8480 = 1.29\).

1 mark for correct option B. Uses the critical angle formula to calculate the refractive index of the adjacent medium.

Resistivity \(\rho\) is related to resistance \(R\) by \(\rho = \frac{R A}{L}\), so the unit of resistivity is \(\Omega \text{ m}\). Resistance is defined as \(R = \frac{V}{I}\), where the unit of current is \(\text{A}\). Potential difference is defined as work done per unit charge, \(V = \frac{W}{Q}\), with units of \(\text{J C}^{-1}\). In base units, work \(W\) is \(\text{kg m}^2 \text{s}^{-2}\) and charge \(Q = I t\) is \(\text{A s}\), so \(V\) has units of \(\text{kg m}^2 \text{s}^{-3} \text{A}^{-1}\). Therefore, resistance \(R\) has units of \(\text{kg m}^2 \text{s}^{-3} \text{A}^{-2}\). Multiplying by meters for resistivity gives \(\text{kg m}^3 \text{s}^{-3} \text{A}^{-2}\).

1 mark for correct option B. Correctly derives the SI base units of potential difference, resistance, and resistivity.

In the third harmonic, the stretched string of length \(L\) contains three half-wavelengths, which means \(L = \frac{3\lambda}{2}\). Rearranging for wavelength gives \(\lambda = \frac{2L}{3}\). The distance between a node and its adjacent antinode is a quarter of a wavelength, \(\frac{\lambda}{4}\). Substituting the expression for wavelength gives the distance as \(\frac{1}{4} \times \frac{2L}{3} = \frac{L}{6}\).

1 mark for correct option B. Relates the string length to the wavelength for the third harmonic and uses the node-to-antinode distance relation.

The current \(I\) is the same in both wires because they are connected in series. Both wires are made of copper, so they have the same number density of conduction electrons \(n\). The charge of an electron \(e\) is constant.

Using the transport equation:
\(I = nAv_e\)

This can be rearranged for the drift velocity:
\(v_e = \frac{I}{nAe}\)

Since \(I\), \(n\), and \(e\) are constant, the drift velocity is inversely proportional to the cross-sectional area:
\(v_e \propto \frac{1}{A}\)

The cross-sectional area of a wire is given by \(A = \frac{\pi d^2}{4}\), which means \(A \propto d^2\). Therefore, \(v_e \propto \frac{1}{d^2}\).

Comparing the two wires:
\(\frac{v_Y}{v_X} = \left(\frac{d_X}{d_Y}\right)^2 = \left(\frac{d}{3d}\right)^2 = \frac{1}{9}\)

Thus, \(v_Y = \frac{1}{9} v_X\).

1 mark for selecting option A.

Incorrect Distractors:
- B: Incorrectly assumes \(v \propto \frac{1}{d}\) rather than \(v \propto \frac{1}{d^2}\).
- C: Incorrectly assumes \(v \propto d\).
- D: Incorrectly assumes \(v \propto d^2\).

According to Einstein's photoelectric equation:
\(hf = \phi + E_k\)

Rearranging to express the photon energy:
\(hf = E_k + \phi\)

When the frequency of the incident light is doubled to \(2f\), the new maximum kinetic energy \(E_{\text{new}}\) is given by:
\(E_{\text{new}} = h(2f) - \phi = 2(hf) - \phi\)

Substituting the expression for \(hf\) into this equation:
\(E_{\text{new}} = 2(E_k + \phi) - \phi\)
\(E_{\text{new}} = 2E_k + 2\phi - \phi\)
\(E_{\text{new}} = 2E_k + \phi\)

1 mark for selecting option B.

Incorrect Distractors:
- A: Incorrectly assumes that doubling the frequency simply doubles the maximum kinetic energy of the electrons without accounting for the constant work function.
- C: Incorrectly subtracts the work function instead of adding it when substituting.
- D: Incorrectly assumes a linear shift without multiplying the initial kinetic energy by 2.

First, calculate the resistance of the wire:
\(R = \rho \frac{l}{A} = 1.7 \times 10^{-8}\ \Omega\text{ m} \times \frac{2.5\text{ m}}{4.5 \times 10^{-7}\text{ m}^2} = 0.0944\ \Omega\)

Next, calculate the current in the wire using Ohm's Law:
\(I = \frac{V}{R} = \frac{1.2\text{ V}}{0.0944\ \Omega} = 12.7\text{ A}\)

Finally, use the transport equation \(I = nAve\) to find the drift velocity:
\(v = \frac{I}{nAe} = \frac{12.7\text{ A}}{8.5 \times 10^{28}\text{ m}^{-3} \times 4.5 \times 10^{-7}\text{ m}^2 \times 1.60 \times 10^{-19}\text{ C}} = 2.08 \times 10^{-3}\text{ m s}^{-1}\)

Rounding to 2 significant figures gives \(2.1 \times 10^{-3}\text{ m s}^{-1}\).

MP1: Use of \(R = \rho \frac{l}{A}\) to find \(R = 0.094\ \Omega\) (or substitution of this into \(I = V/R\)) [1M]
MP2: Use of \(I = nAve\) with calculated current \(I = 12.7\text{ A}\) [1M]
MP3: Correct final value of drift velocity in range \(2.08 \times 10^{-3}\text{ m s}^{-1}\) to \(2.1 \times 10^{-3}\text{ m s}^{-1}\) [1M]

First, calculate the energy of an incident photon:
\(E = hf = 6.63 \times 10^{-34}\text{ J s} \times 1.2 \times 10^{15}\text{ Hz} = 7.96 \times 10^{-19}\text{ J}\)

Next, convert the work function of zinc into joules:
\(\Phi = 4.3\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 6.88 \times 10^{-19}\text{ J}\)

Since the photon energy \(E\) (\(7.96 \times 10^{-19}\text{ J}\)) is greater than the work function \(\Phi\) (\(6.88 \times 10^{-19}\text{ J}\)), photoelectric emission does occur.

Finally, calculate the maximum kinetic energy of the photoelectrons using Einstein's photoelectric equation:
\(E_{k,\text{max}} = hf - \Phi = 7.96 \times 10^{-19}\text{ J} - 6.88 \times 10^{-19}\text{ J} = 1.08 \times 10^{-19}\text{ J}\)

Rounding to 2 significant figures gives \(1.1 \times 10^{-19}\text{ J\) (or \(1.08 \times 10^{-19}\text{ J}\) to 3 s.f.).

MP1: Calculation of photon energy in J (\(7.96 \times 10^{-19}\text{ J}\)) OR work function in J (\(6.88 \times 10^{-19}\text{ J}\)) [1M]
MP2: Comparison showing photon energy is greater than work function AND statement that emission occurs [1M]
MP3: Correct calculation of maximum kinetic energy as \(1.1 \times 10^{-19}\text{ J}\) (accept \(1.08 \times 10^{-19}\text{ J}\)) [1M]

For a wire fixed at both ends vibrating at its third harmonic, there are three loops (half-wavelengths) along its length \(L\).
Therefore, the relationship between the length of the wire and the wavelength \(\lambda\) is:
\(L = \frac{3}{2}\lambda\)

Rearranging to find the wavelength:
\(\lambda = \frac{2}{3}L = \frac{2}{3} \times 1.20\text{ m} = 0.80\text{ m}\)

Use the wave equation to find the frequency:
\(v = f\lambda \implies f = \frac{v}{\lambda}\)
\(f = \frac{180\text{ m s}^{-1}}{0.80\text{ m}} = 225\text{ Hz}\)

MP1: Recall/use of relationship \(L = 1.5\lambda\) for the third harmonic [1M]
MP2: Calculation of wavelength \(\lambda = 0.80\text{ m}\) [1M]
MP3: Correct frequency calculation \(f = 225\text{ Hz}\) (accept \(230\text{ Hz}\)) [1M]

(a) Using the formula for the critical angle:
\(\sin\theta_c = \frac{n_2}{n_1} = \frac{1.44}{1.48}\)
\(\theta_c = \arcsin(0.9730) = 76.7^\circ\) (or \(1.34\text{ rad}\))

(b) The speed of light in the core is:
\(v = \frac{c}{n_1} = \frac{3.00 \times 10^8\text{ m s}^{-1}}{1.48} = 2.027 \times 10^8\text{ m s}^{-1}\)

The time taken for a ray travelling straight along the axis of length \(L = 2500\text{ m}\) is:
\(t_{\text{min}} = \frac{L}{v} = \frac{2500\text{ m}}{2.027 \times 10^8\text{ m s}^{-1}} = 1.233 \times 10^{-5}\text{ s}\)

The longest path is taken by the ray reflecting at the critical angle \(\theta_c\). The path length is:
\(L' = \frac{L}{\sin\theta_c} = L \times \frac{n_1}{n_2} = 2500\text{ m} \times \frac{1.48}{1.44} = 2569.4\text{ m}\)

The time taken for this longest path is:
\(t_{\text{max}} = \frac{L'}{v} = \frac{2569.4\text{ m}}{2.027 \times 10^8\text{ m s}^{-1}} = 1.268 \times 10^{-5}\text{ s}\)

The maximum time delay is:
\(\Delta t = t_{\text{max}} - t_{\text{min}} = 1.268 \times 10^{-5}\text{ s} - 1.233 \times 10^{-5}\text{ s} = 3.5 \times 10^{-7}\text{ s}\)

(a)
- Use of \(\sin \theta_c = \frac{n_2}{n_1}\) [1 mark]
- Correct value for \(\theta_c = 76.7^\circ\) (Accept range \(76.5^\circ\) to \(77.0^\circ\)) [1 mark]

(b)
- Calculates speed of light in core \(v = 2.03 \times 10^8\text{ m s}^{-1}\) [1 mark]
- Calculates minimum time \(t_{\text{min}} = 1.23 \times 10^{-5}\text{ s}\) [1 mark]
- Determines longest path \(L' = 2570\text{ m}\) [1 mark]
- Calculates maximum time \(t_{\text{max}} = 1.27 \times 10^{-5}\text{ s}\) [1 mark]
- Calculates time delay \(\Delta t = 3.5 \times 10^{-7}\text{ s}\) (Accept answers in range \(3.4 \times 10^{-7}\text{ s}\) to \(3.5 \times 10^{-7}\text{ s}\)) [1 mark]

(a) First, find the current \(I\) in the circuit:
\(I = \frac{V}{R} = \frac{0.48\text{ V}}{4.0\ \Omega} = 0.12\text{ A}\)

Using the e.m.f. equation:
\(E = V + Ir\)
\(0.62\text{ V} = 0.48\text{ V} + (0.12\text{ A}) \times r\)
\(0.14\text{ V} = 0.12\text{ A} \times r\)
\(r = 1.17\ \Omega\)

(b) Under maximum power conditions, \(R = r = 1.17\ \Omega\).
The total resistance in the circuit is:
\(R_{\text{total}} = R + r = 1.17\ \Omega + 1.17\ \Omega = 2.34\ \Omega\)

The current in the circuit is:
\(I = \frac{E}{R_{\text{total}}} = \frac{0.62\text{ V}}{2.34\ \Omega} = 0.265\text{ A}\)

The maximum power delivered to \(R\) is:
\(P = I^2 R = (0.265\text{ A})^2 \times 1.17\ \Omega = 0.082\text{ W}\) (or \(82\text{ mW}\))

The efficiency \(\eta\) of the cell is:
\(\eta = \frac{\text{Useful power output}}{\text{Total power input}} = \frac{I^2 R}{I^2(R+r)} = \frac{R}{R+r} = \frac{1.17}{2.34} = 50\%\)

(a)
- Calculates circuit current \(I = 0.12\text{ A}\) [1 mark]
- Uses \(E = V + Ir\) [1 mark]
- Correctly calculates \(r = 1.17\ \Omega\) (Accept range \(1.1\ \Omega\) to \(1.2\ \Omega\)) [1 mark]

(b)
- Identifies \(R_{\text{total}} = 2r = 2.34\ \Omega\) [1 mark]
- Calculates new current \(I = 0.265\text{ A}\) [1 mark]
- Calculates power \(P = 0.082\text{ W}\) (Accept answers in range \(0.080\text{ W}\) to \(0.083\text{ W}\)) [1 mark]
- Correctly identifies efficiency is \(50\%\) [1 mark]

(a) First, calculate the energy \(E\) of the incident photons:
\(E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34}\text{ J s})(3.00 \times 10^8\text{ m s}^{-1})}{240 \times 10^{-9}\text{ m}} = 8.29 \times 10^{-19}\text{ J}\)

Convert the work function \(\Phi\) from \(\text{eV}\) to joules:
\(\Phi = 3.40\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 5.44 \times 10^{-19}\text{ J}\)

Apply Einstein's photoelectric equation:
\(E_k = hf - \Phi = 8.29 \times 10^{-19}\text{ J} - 5.44 \times 10^{-19}\text{ J} = 2.85 \times 10^{-19}\text{ J}\)

(b) The de Broglie wavelength \(\lambda_{\text{dB}}\) is given by:
\(\lambda_{\text{dB}} = \frac{h}{p}\)

Using \(E_k = \frac{p^2}{2m}\), the momentum \(p\) of the electron is:
\(p = \sqrt{2 m_e E_k} = \sqrt{2 \times (9.11 \times 10^{-31}\text{ kg}) \times (2.85 \times 10^{-19}\text{ J})} = 7.21 \times 10^{-25}\text{ kg m s}^{-1}\)

Thus, the de Broglie wavelength is:
\(\lambda_{\text{dB}} = \frac{6.63 \times 10^{-34}\text{ J s}}{7.21 \times 10^{-25}\text{ kg m s}^{-1}} = 9.20 \times 10^{-10}\text{ m}\) (or \(0.92\text{ nm}\))

(a)
- Uses \(E = \frac{hc}{\lambda}\) to find photon energy \(E = 8.29 \times 10^{-19}\text{ J}\) [1 mark]
- Converts \(\text{eV}\) to Joules (\(\Phi = 5.44 \times 10^{-19}\text{ J}\)) [1 mark]
- Recalls Einstein's photoelectric equation \(E_k = hf - \Phi\) [1 mark]
- Obtains maximum kinetic energy \(E_k = 2.85 \times 10^{-19}\text{ J}\) (accept range \(2.8 \times 10^{-19}\text{ J}\) to \(2.9 \times 10^{-19}\text{ J}\)) [1 mark]

(b)
- Uses relationship \(p = \sqrt{2mE_k}\) or calculates electron speed \(v = \sqrt{2E_k/m} = 7.91 \times 10^5\text{ m s}^{-1}\) [1 mark]
- Uses de Broglie equation \(\lambda = \frac{h}{p}\) [1 mark]
- Obtains wavelength \(\lambda_{\text{dB}} = 9.20 \times 10^{-10}\text{ m}\) (accept range \(9.10 \times 10^{-10}\text{ m}\) to \(9.30 \times 10^{-10}\text{ m}\)) [1 mark]

(a) First, calculate the cross-sectional area \(A\) of the wire:
\(A = \frac{\pi d^2}{4} = \frac{\pi (0.45 \times 10^{-3}\text{ m})^2}{4} = 1.59 \times 10^{-7}\text{ m}^2\)

Now calculate the resistance \(R\) using the resistivity formula:
\(R = \frac{\rho L}{A} = \frac{(1.7 \times 10^{-8}\ \Omega\text{ m})(3.5\text{ m})}{1.59 \times 10^{-7}\text{ m}^2} = 0.374\ \Omega\)

(b) Use the transport equation for electric current:
\(I = n A v q\)

Rearranging for drift velocity \(v\):
\(v = \frac{I}{n A e}\)

Substitute the known values, with the charge of an electron \(e = 1.60 \times 10^{-19}\text{ C}\):
\(v = \frac{1.8\text{ A}}{(8.5 \times 10^{28}\text{ m}^{-3})(1.59 \times 10^{-7}\text{ m}^2)(1.60 \times 10^{-19}\text{ C})}\)
\(v = \frac{1.8}{2162.4} = 8.32 \times 10^{-4}\text{ m s}^{-1}\)

(a)
- Calculates cross-sectional area \(A = 1.59 \times 10^{-7}\text{ m}^2\) [1 mark]
- Uses \(R = \frac{\rho L}{A}\) [1 mark]
- Correctly obtains resistance \(R = 0.37\ \Omega\) (accept range \(0.37\ \Omega\) to \(0.38\ \Omega\)) [1 mark]

(b)
- Recalls transport equation \(I = n A v q\) [1 mark]
- Substitutes electronic charge \(e = 1.60 \times 10^{-19}\text{ C}\) [1 mark]
- Correct rearrangement for \(v\) [1 mark]
- Calculates drift velocity \(v = 8.3 \times 10^{-4}\text{ m s}^{-1}\) (accept range \(8.2 \times 10^{-4}\text{ m s}^{-1}\) to \(8.4 \times 10^{-4}\text{ m s}^{-1}\)) [1 mark]

(a) First, find the grating spacing \(d\):
\(d = \frac{1 \times 10^{-3}\text{ m}}{500} = 2.00 \times 10^{-6}\text{ m}\)

Use the grating equation \(d \sin \theta = n \lambda\) for the second-order maximum (\(n = 2\)):
For \(\lambda_1 = 589.0\text{ nm}\):
\(\sin \theta_1 = \frac{2 \times 589.0 \times 10^{-9}\text{ m}}{2.00 \times 10^{-6}\text{ m}} = 0.589\)
\(\theta_1 = \arcsin(0.589) = 36.085^\circ\)

For \(\lambda_2 = 589.6\text{ nm}\):
\(\sin \theta_2 = \frac{2 \times 589.6 \times 10^{-9}\text{ m}}{2.00 \times 10^{-6}\text{ m}} = 0.5896\)
\(\theta_2 = \arcsin(0.5896) = 36.129^\circ\)

The angular separation \(\Delta \theta\) is:
\(\Delta \theta = \theta_2 - \theta_1 = 36.129^\circ - 36.085^\circ = 0.044^\circ\)

(b) The distance \(y\) from the central maximum on the screen is given by \(y = D \tan \theta\), where \(D = 2.20\text{ m\}}.
For \)\lambda_1\):
\(y_1 = 2.20\text{ m} \times \tan(36.085^\circ) = 1.6035\text{ m}\)

For \(\lambda_2\):
\(y_2 = 2.20\text{ m} \times \tan(36.129^\circ) = 1.6061\text{ m}\)

The linear separation \(\Delta y\) is:
\(\Delta y = y_2 - y_1 = 1.6061\text{ m} - 1.6035\text{ m} = 0.0026\text{ m}\) (or \(2.6\text{ mm}\))

(a)
- Calculates grating spacing \(d = 2.00 \times 10^{-6}\text{ m}\) [1 mark]
- Recalls grating equation \(d \sin \theta = n \lambda\) [1 mark]
- Calculates both angles \(\theta_1 = 36.09^\circ\) and \(\theta_2 = 36.13^\circ\) [1 mark]
- Calculates difference \(\Delta \theta = 0.044^\circ\) (or \(7.7 \times 10^{-4}\text{ rad}\)) [1 mark]

(b)
- Recalls trigonometric relation \(y = D \tan \theta\) [1 mark]
- Calculates positions \(y_1 = 1.604\text{ m}\) and \(y_2 = 1.606\text{ m}\) [1 mark]
- Calculates linear separation \(\Delta y = 2.6\text{ mm}\) (or \(2.6 \times 10^{-3}\text{ m}\)) [1 mark]

(a) Using the potential divider equation:
\(V_{\text{out}} = V_{\text{in}} \times \frac{R_T}{R_1 + R_T}\)

Substitute the given values:
\(6.0\text{ V} = 9.0\text{ V} \times \frac{R_T}{1800\ \Omega + R_T}\)
\(\frac{6.0}{9.0} = \frac{R_T}{1800 + R_T}\)
\(\frac{2}{3} = \frac{R_T}{1800 + R_T}\)
\(2(1800 + R_T) = 3 R_T\)
\(3600 + 2 R_T = 3 R_T \implies R_T = 3600\ \Omega = 3.6\ \text{k}\Omega\)

(b) At \(60^\circ\text{C}\), \(R_T = 1.2\ \text{k}\Omega = 1200\ \Omega\).
The new output voltage is:
\(V_{\text{out}} = 9.0\text{ V} \times \frac{1200\ \Omega}{1800\ \Omega + 1200\ \Omega} = 9.0\text{ V} \times \frac{1200}{3000} = 3.6\text{ V}\)

The electrical power \(P\) dissipated by the thermistor is:
\(P = \frac{V_{\text{out}}^2}{R_T} = \frac{(3.6\text{ V})^2}{1200\ \Omega} = \frac{12.96}{1200} = 0.0108\text{ W}\) (or \(10.8\text{ mW}\))

(a)
- Recalls potential divider relationship [1 mark]
- Substitutes values into the expression [1 mark]
- Correctly calculates \(R_T = 3.6\ \text{k}\Omega\) (or \(3600\ \Omega\)) [1 mark]

(b)
- Recalculates new output voltage \(V_{\text{out}} = 3.6\text{ V}\) [2 marks] (Award 1 mark for correct method, 1 mark for correct value)
- Recalls power equation \(P = \frac{V^2}{R}\) or \(P = I^2 R\) or \(P = IV\) [1 mark]
- Calculates power \(P = 1.1 \times 10^{-2}\text{ W}\) (accept range \(1.05 \times 10^{-2}\text{ W}\) to \(1.1 \times 10^{-2}\text{ W}\) or \(10.8\text{ mW}\)) [1 mark]

(a)
- When a potential difference is applied, electrons gain electrical energy and are promoted across the bandgap to a higher energy level (conduction band).
- These electrons then transition back down to a lower energy level (recombine with holes in the valence band).
- During this transition, they lose a specific packet of energy, which is emitted as a single photon of light of a specific frequency \(f\) and wavelength \(\lambda\), where the photon energy \(E = hf = \frac{hc}{\lambda}\).

(b)
- The relationship is given by the equation: \(eV_0 = \frac{hc}{\lambda}\) which can be rewritten as \(V_0 = \left(\frac{hc}{e}\right)\frac{1}{\lambda}\).
- Comparing this to the equation of a straight line, \(y = mx + c\), a plot of \(V_0\) on the y-axis against \(1/\lambda\) on the x-axis yields a straight line through the origin.
- The gradient of this line is \(m = \frac{hc}{e}\). Hence, Planck's constant can be calculated as \(h = \frac{\text{gradient} \times e}{c}\).

(c)
- From part (b), \(h = \frac{\text{gradient} \times e}{c}\).
- Substituting the given values: \(h = \frac{1.28 \times 10^{-6}\text{ V m} \times 1.60 \times 10^{-19}\text{ C}}{3.00 \times 10^8\text{ m s}^{-1}} = 6.83 \times 10^{-34}\text{ J s}\).

(d)
- Place a sensitive ammeter (microammeter or milliammeter) in series with the LED to measure the current.
- Vary the potential difference, record the current, and plot a graph of current \(I\) against voltage \(V\). Extrapolate the linear portion of the curve back to zero current to find the precise threshold voltage \(V_0\).

(a)
- MP1: Electrons gain energy from the potential difference and move to a higher energy level / conduction band (1)
- MP2: Electrons drop back to a lower energy level / recombine with holes (1)
- MP3: Energy is released as a photon where \(E = hf\) or \(E = \frac{hc}{\lambda}\) (1)

(b)
- MP1: State equation: \(V_0 = \frac{hc}{e\lambda}\) (1)
- MP2: State that a graph of \(V_0\) against \(1/\lambda\) is linear with a gradient of \(\frac{hc}{e}\) (1)
- MP3: State that \(h = \frac{\text{gradient} \times e}{c}\) (1)

(c)
- MP1: Correct substitution of values into the rearranged equation (1)
- MP2: Correct calculation of \(h = 6.83 \times 10^{-34}\text{ J s}\) (accept \(6.8 \times 10^{-34}\text{ J s}\)) (1)

(d)
- MP1: Measure current through the LED using a sensitive ammeter / milliammeter / microammeter (1)
- MP2: Plot current against potential difference and extrapolate the straight-line section to the voltage axis where \(I = 0\) to find \(V_0\) (1)

(a)
- As temperature increases, the resistance of the NTC thermistor decreases.
- This decreases the total resistance of the circuit, causing the current in the circuit to increase. Since \(V_{\text{out}} = IR\) and \(R\) is constant, the potential difference across the fixed resistor increases.

(b)
- The total resistance of the potential divider circuit is \(R_{\text{total}} = R + R_{\text{th}} = 1200\ \Omega + 850\ \Omega = 2050\ \Omega\).
- Using the potential divider formula: \(V_{\text{out}} = V_{\text{in}} \times \frac{R}{R_{\text{total}}}\).
- \(V_{\text{out}} = 9.0\text{ V} \times \frac{1200\ \Omega}{2050\ \Omega} = 5.27\text{ V}\) (which rounds to \(5.3\text{ V}\)).

(c)
- In a semiconductor (such as a thermistor), an increase in temperature provides thermal energy to the atoms.
- This causes valence electrons to gain enough energy to break free and move into the conduction band, increasing the number density \(n\) of charge carriers.
- Although the lattice ions vibrate more, causing more frequent collisions with charge carriers, the massive increase in the charge carrier density dominates, resulting in a net decrease in resistance.

(d)
- At a lower temperature, the resistance of the thermistor \(R_{\text{th}}\) is higher.
- For \(V_{\text{out}}\) to remain at \(5.3\text{ V}\), the ratio of the fixed resistance \(R\) to the total resistance must stay the same, meaning the ratio \(R/R_{\text{th}}\) must remain constant. Therefore, the resistance of the fixed resistor must be increased.

(a)
- MP1: Resistance of the thermistor decreases as temperature increases (1)
- MP2: Causes circuit current to increase / less potential difference is dropped across the thermistor, so more potential difference is dropped across the fixed resistor (1)

(b)
- MP1: Calculate total resistance: \(R_{\text{total}} = 2050\ \Omega\) (1)
- MP2: Use correct potential divider formula with substitution (1)
- MP3: Calculate \(V_{\text{out}} = 5.27\text{ V}\) (or \(5.3\text{ V}\)) (1)

(c)
- MP1: Higher temperature releases more charge carriers / electrons into the conduction band (number density \(n\) increases) (1)
- MP2: Larger number of charge carriers reduces resistance (1)
- MP3: Recognise that this effect of increasing \(n\) dominates over the effect of increased lattice vibrations / collisions (1)

(d)
- MP1: Identify that at lower temperature, the thermistor resistance \(R_{\text{th}}\) increases (1)
- MP2: To keep the ratio of resistances the same (to maintain \(V_{\text{out}} = 5.3\text{ V}\)), the resistance of the fixed resistor must be increased (1)

(a) Criticism: The reading of 0.44 mm is an anomaly and should be discarded. The remaining four readings are consistent. Mean diameter calculation: (0.38 + 0.37 + 0.38 + 0.38) / 4 = 0.3775 mm. Rounded to 2 d.p. to match micrometer precision, mean = 0.38 mm. Uncertainty is half-range: (0.38 - 0.37)/2 = 0.005 mm (or 0.01 mm as it matches resolution). (b) Plot a graph of resistance \(R\) on the y-axis against length \(L\) on the x-axis. Since \(R = \frac{\rho L}{A}\), the gradient \(m = \frac{\rho}{A}\), so \(\rho = m \times A\), where \(A = \frac{\pi d^2}{4}\). Unit of gradient is \(\Omega\text{ m}^{-1}\), and unit of area is \(\text{m}^2\). Multiplying gradient by area gives \(\Omega\text{ m}^{-1} \times \text{m}^2 = \Omega\text{ m}\). (c) The 20 \(\Omega\) range displays to two decimal places, so '2 digits' equals 0.02 \(\Omega\). Absolute uncertainty = \(0.8\%\text{ of } 4.42\ \Omega + 0.02\ \Omega = 0.03536 + 0.02 = 0.05536\ \Omega\). Percentage uncertainty = \((0.05536 / 4.42) \times 100\% = 1.25\% \approx 1.3\%\). (d) Replace crocodile clips with a metal jockey or sharp knife-edge pointer to make electrical contact at precise lengths, which reduces contact resistance and increases precision in length measurement.

(a) MP1: Identify 0.44 mm as anomalous and state it should be excluded. MP2: Correctly calculate the mean diameter as 0.38 mm (or 0.378 mm). MP3: Calculate absolute uncertainty as half the range (0.005 mm) or recognize the resolution limit of the micrometer (0.01 mm). MP4: Express final result correctly with unit (e.g., 0.38 \(\pm\) 0.01 mm or 0.38 \(\pm\) 0.005 mm). (b) MP1: Plot graph of R against L and state that the gradient of this graph represents \(\rho/A\). MP2: Show that \(\rho = \text{gradient} \times \text{area}\) where \(A = \pi d^2 / 4\). MP3: Show unit of gradient is \(\Omega\text{ m}^{-1}\) and unit of area is \(\text{m}^2\). MP4: Deduce unit of resistivity is \(\Omega\text{ m}\). (c) MP1: Correctly interpret 2 digits as 0.02 \(\Omega\) to calculate total absolute uncertainty of 0.055 \(\Omega\). MP2: Calculate percentage uncertainty as 1.25\% or 1.3\%. (d) MP1: Suggest using a jockey or sharp-edged knife-pointer. MP2: Explain that this provides a smaller, more reproducible contact area, minimising contact resistance (or improves precision of length measurement).

(a) Mean time \(t = (0.412 + 0.419 + 0.415 + 0.411 + 0.413) / 5 = 0.414\text{ s}\). Range of time = 0.419 - 0.411 = 0.008 s. Uncertainty in time = half-range = 0.004 s. Percentage uncertainty in \(t = (0.004 / 0.414) \times 100\% = 0.966\% \approx 0.97\%\) (accept 1.0\%). (b) Using \(h = \frac{1}{2}gt^2\), we find \(g = \frac{2h}{t^2}\). Substituting values: \(g = \frac{2 \times 0.850}{(0.414)^2} = 9.918\text{ m s}^{-2} \approx 9.9\text{ m s}^{-2}\) (which is approximately 10 \(\text{m s}^{-2}\)). Percentage uncertainty in \(h = (0.002 / 0.850) \times 100\% = 0.235\%\). Percentage uncertainty in \(g = \%\Delta h + 2 \times \%\Delta t = 0.235\% + 2 \times 0.966\% = 2.167\% \approx 2.2\%\). (c) (i) Since the timer starts before the ball is released, the recorded time of fall \(t\) will be larger than the actual time of fall. (ii) Because \(g \propto 1/t^2\), a larger measured value of \(t\) will lead to a smaller calculated value of \(g\) (it will underestimate \(g\)). (d) The student could use a mechanical release mechanism (such as a spring-loaded trapdoor switch that breaks the timer circuit at the exact moment of release), or they could vary the height \(h\), plot a graph of \(h\) against \(t^2\), and find \(g\) from the gradient (since any constant delay in release will only cause a non-zero intercept on the time axis but won't affect the gradient of the line).

(a) MP1: Calculate mean time as 0.414 s. MP2: Determine the absolute uncertainty in time as half the range (0.004 s). MP3: Calculate percentage uncertainty in time as 0.97\% (or 1\% if rounded to 1 s.f.). (b) MP1: Use \(g = 2h/t^2\) to show \(g \approx 9.9\text{ m s}^{-2}\). MP2: Calculate percentage uncertainty in \(h\) as 0.24\% (or 0.235\%). MP3: Use formula \(\%\Delta g = \%\Delta h + 2\%\Delta t\). MP4: Obtain percentage uncertainty in \(g\) as 2.2\% (allow 2.0\% - 2.3\% depending on rounding of \(t\) uncertainty). (c) MP1: State that the measured time \(t\) is longer than the actual fall time. MP2: State that the calculated \(g\) is smaller than the true value. MP3: Explain that since \(g = 2h/t^2\), a larger denominator leads to a smaller overall quotient. (d) MP1: Suggest a mechanical release system (or a magnetic release with a fast-acting coil/demagnetiser circuit). MP2: Alternatively, suggest plotting a graph of \(h\) against \(t^2\) and using the gradient, explaining that a constant systematic delay only creates an intercept on the axis and does not alter the gradient.

1. Circuit Diagram: The diagram must show a DC power supply or cell, connected to an ammeter in series, and a voltmeter connected in parallel across the variable length of the test wire. A variable resistor or switch should also be included. Connections to the wire should be clearly marked with a movable contact (such as a flying lead or crocodile clip) so that different lengths of the wire can be tested. 2. Experimental Procedure: Adjust the position of the contact to select a length of wire. Close the switch, measure the current \(I\) and voltage \(V\), then immediately open the switch to minimize heating. Repeat this for at least five different lengths \(L\) (measured using a metre rule) ranging from 0.1 m to 1.0 m. Keep the current small to prevent temperature changes that would alter the resistance. 3. Cross-sectional Area: Use a micrometer screw gauge to measure the diameter \(d\) of the wire. Take measurements at at least five different positions along the wire and at different perpendicular orientations to account for any non-uniformity. Calculate the mean diameter and use the formula \(A = \frac{\pi d^2}{4}\) to find the area. 4. Analysis: Calculate the resistance \(R = \frac{V}{I}\) for each length. Plot a graph of Resistance \(R\) on the y-axis against length \(L\) on the x-axis. The graph should be a straight line passing through the origin. Since \(R = \frac{\rho L}{A}\), the gradient of this graph represents \(\frac{\rho}{A}\). Therefore, the resistivity of constantan can be calculated as \(\rho = \text{gradient} \times A\).

1. Diagram (3 marks): [1] Correct symbols for power supply, ammeter, and voltmeter. [1] Ammeter in series with the test wire, voltmeter in parallel across the variable length of the wire. [1] Flying lead or sliding contact shown clearly to vary the length of the wire in the circuit. 2. Procedure (3 marks): [1] Measurement of length \(L\) using a metre rule. [1] Measurement of current \(I\) and potential difference \(V\) for different lengths of wire. [1] Switch off the circuit between readings or use a variable resistor to keep the current low to minimize temperature rise/heating. 3. Cross-sectional Area (3 marks): [1] Use a micrometer screw gauge to measure the wire's diameter \(d\). [1] Measure diameter at multiple positions and orientations, then calculate the average. [1] Correct equation for cross-sectional area \(A = \frac{\pi d^2}{4}\). 4. Analysis (4 marks): [1] Calculate resistance \(R = \frac{V}{I}\) for each length. [1] Plot a graph of \(R\) against \(L\). [1] Sketch graph shows a straight line through the origin. [1] State that \(\rho = \text{gradient} \times A\) (accept gradient = \(\frac{\rho}{A}\)).

1. Experimental Apparatus: The diagram should show a long test wire fixed at one end to a rigid support (or clamped to a bench with a pulley system). A weight hanger is attached to the free end to apply known loads. A marker is fixed to the wire, and a metre rule is placed next to it to measure the movement of the marker (extension). Alternatively, Searle's apparatus with a vernier scale or a traveling microscope can be shown. 2. Measurements and Instruments: Measure the original length \(L_0\) of the test wire from the clamped end to the marker using a metre rule. Measure the diameter \(d\) of the wire at several points and orientations using a micrometer screw gauge, and find the mean diameter. For each hanging mass \(m\), calculate the tensile force \(F = mg\). Record the position of the marker and subtract the initial position to find the extension \(\Delta x\). 3. Safety and Accuracy: Wear safety goggles to protect the eyes in case the wire breaks. Place a soft tray or box filled with foam under the weights to catch them safely if the wire snaps. To improve accuracy of the extension measurement, use a very long wire (e.g., over 2.0 m) to maximize the extension, and read the scale at eye level to avoid parallax error. 4. Analysis: Calculate the cross-sectional area of the wire \(A = \frac{\pi d^2}{4}\). Calculate stress \(\sigma = \frac{F}{A}\) and strain \(\epsilon = \frac{\Delta x}{L_0}\). Plot a graph of stress against strain. The gradient of the linear region of this graph is the Young modulus \(E\). Alternatively, plot a graph of Force \(F\) against extension \(\Delta x\). The gradient of the linear section is \(\frac{E A}{L_0}\), from which \(E = \text{gradient} \times \frac{L_0}{A}\).

1. Diagram (3 marks): [1] Rigorous support/clamp and pulley arrangement or vertical clamp shown. [1] Hanging masses at the end of the wire. [1] Metre rule and a marker on the wire (or traveling microscope) to measure extension. 2. Measurements and Instruments (4 marks): [1] Original length \(L_0\) measured with a metre rule. [1] Diameter \(d\) measured with a micrometer. [1] Diameter measured at multiple positions and orientations to find a mean. [1] Extension \(\Delta x\) measured as change in position of the marker using a rule/microscope. 3. Safety and Accuracy (2 marks): [1] Wear safety goggles (to protect eyes from snapping wire) OR place a padded box beneath weights. [1] Use a long wire to obtain a large, measurable extension OR use a marker/vernier to reduce parallax/scale reading uncertainty. 4. Analysis (4 marks): [1] Calculate cross-sectional area \(A = \frac{\pi d^2}{4}\). [1] Plot a graph of force \(F\) against extension \(\Delta x\) (or stress against strain). [1] Draw a straight line through the linear region of the data points. [1] State how to find the Young modulus from the gradient (e.g. \(E = \text{gradient} \times \frac{L_0}{A}\) for \(F\)-\(\Delta x\) graph, or \(E = \text{gradient}\) for stress-strain graph).