2026 Edexcel IAS-Level Physics (XPH11) 模擬試題連答案詳解

The resultant velocity is the vector sum of the two perpendicular velocities. The magnitude of the resultant velocity is calculated using Pythagoras' theorem: \(v = \sqrt{3.0^2 + 4.0^2} = 5.0\text{ m s}^{-1}\). The angle \(\theta\) East of North is found using trigonometry: \(\tan\theta = \frac{4.0}{3.0}\), which gives \(\theta \approx 53^\circ\).

[1] Correctly identifies Option A as the only option with both the correct magnitude of 5.0 m s^-1 and correct angle of 53 degrees East of North.

Using the equation of motion \(v = u - gt\), at maximum height \(v = 0\), so \(t = \frac{u}{g}\). Doubling \(u\) doubles the time taken to \(2t\). Using \(v^2 = u^2 - 2gh\), at maximum height \(v = 0\), so \(h = \frac{u^2}{2g}\). Doubling \(u\) increases the maximum height by a factor of \(2^2 = 4\), giving \(4h\).

[1] B is the correct answer based on the quadratic relationship of initial velocity to height and linear relationship to time.

Resolving forces perpendicular to the inclined plane: the normal contact force \(N\) acts perpendicular to the surface of the ramp, and the component of the block's weight perpendicular to the ramp is \(mg \cos\theta\). Since there is no acceleration perpendicular to the ramp, these forces must balance, so \(N = mg \cos\theta\).

[1] B is correct because the component of weight perpendicular to the ramp is balanced by the normal contact force.

First, find the useful work done: \(W = Fd = 50\text{ N} \times 3.0\text{ m} = 150\text{ J}\). The useful power output of the motor is \(P_{\text{out}} = \frac{W}{t} = \frac{150\text{ J}}{5.0\text{ s}} = 30\text{ W}\). Since the motor is \(60\%\) efficient, \(\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \implies 0.60 = \frac{30\text{ W}}{P_{\text{in}}}\). Therefore, the power input is \(P_{\text{in}} = \frac{30}{0.60} = 50\text{ W}\).

[1] C is correct by calculating useful power output (30 W) and dividing by efficiency (0.60).

Using conservation of momentum: \(m_1 u_1 + m_2 u_2 = (m_1 + m_2)v \implies (2.0 \times 4.0) + (3.0 \times 0) = (2.0 + 3.0)v \implies 8.0 = 5.0v \implies v = 1.6\text{ m s}^{-1}\). Because the two trolleys couple together, kinetic energy is not conserved (some is converted to thermal energy and sound), making the collision inelastic.

[1] B is correct because momentum conservation yields 1.6 m s^-1, and coupling always indicates an inelastic collision.

The extension is given by \(\Delta L = \frac{F L}{A E}\). Since both wires are made of the same material, the Young Modulus \(E\) is the same, and the force \(F\) is also the same. The cross-sectional area \(A\) is proportional to the square of the diameter \(d^2\), so \(\Delta L \propto \frac{L}{d^2}\). Thus, the ratio is \(\frac{\Delta L_X}{\Delta L_Y} = \frac{L_X}{L_Y} \times \left(\frac{d_Y}{d_X}\right)^2 = 2 \times \left(\frac{1}{2}\right)^2 = 0.5\).

[1] A is correct because the doubling of length increases extension by 2, but doubling diameter increases area by 4, reducing extension by 1/4, resulting in a net factor of 0.5.

At terminal velocity, drag force equals weight: \(6 \pi \eta r v = mg\). Since mass \(m = \rho V = \rho \frac{4}{3} \pi r^3\), the weight is proportional to \(r^3\). Therefore, \(6 \pi \eta r v \propto r^3 \implies v \propto r^2\). If the radius is doubled to \(2r\), the terminal velocity is multiplied by \(2^2 = 4\), which is \(4v\).

[1] C is correct because terminal velocity is proportional to the square of the radius when upthrust is negligible.

When a ductile material like copper is stretched beyond its elastic limit, it undergoes plastic deformation. When the load is removed, the atoms slide back elastically but do not return to their original positions completely. The unloading line on a force-extension or stress-strain graph is straight and parallel to the initial Hooke's Law region, resulting in a permanent extension (plastic strain).

[1] C is correct because plastic deformation results in a parallel unloading line and a permanent extension.

According to Newton's second law of motion, the rate of change of momentum of an object is equal to the net resultant force acting on it:

\(F = \frac{\Delta p}{\Delta t}\)

Throughout the projectile's flight, the only force acting on it is the gravitational force (its weight), which acts vertically downwards with a constant magnitude of \(mg\).

Therefore, at any point during its flight, including at the highest point, the rate of change of momentum of the projectile is equal to \(mg\).

C is the correct answer.
- 1 mark for identifying that the rate of change of momentum is equal to the net force acting on the projectile, which is the constant weight \(mg\).

The Young modulus \(E\) of a material is given by:

\(E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{\Delta L / L} = \frac{F L}{A \Delta L}\)

Rearranging this expression for the extension \(\Delta L\):

\(\Delta L = \frac{F L}{A E}\)

The cross-sectional area \(A\) of a wire with diameter \(d\) is:

\(A = \frac{\pi d^2}{4}\)

Substituting \(A\) into the equation for extension gives:

\(\Delta L = \frac{4 F L}{\pi d^2 E}\)

Since both wires are made of the same material, they have the same Young modulus \(E\). They are also subjected to the same tensile force \(F\).

Therefore, the extension is proportional to \(\frac{L}{d^2}\):

\(\Delta L \propto \frac{L}{d^2}\)

For Wire X:

\(\Delta L_X \propto \frac{L}{d^2}\)

For Wire Y:

\(\Delta L_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2}\)

Comparing the two extensions:

\(\frac{\Delta L_X}{\Delta L_Y} = \frac{L / d^2}{L / (2d^2)} = 2\)

C is the correct answer.
- 1 mark for expressing extension in terms of length and diameter, and correctly determining that the extension of Wire X is twice the extension of Wire Y.

(a) Using the equation of motion vertically: \(s = ut + 0.5 a t^2\). Since the initial vertical velocity is zero: \(45.0 = 0.5 \times 9.81 \times t^2\). Rearranging gives \(t^2 = 9.17\), hence \(t = 3.03\text{ s}\), which is approximately \(3\text{ s}\). (b) The horizontal distance is calculated using \(x = u_x \times t = 15.0 \times 3.03 = 45.5\text{ m}\). The vertical component of velocity just before impact is \(v_y = u_y + at = 0 + (9.81 \times 3.03) = 29.7\text{ m s}^{-1}\). The horizontal component of velocity is constant: \(v_x = 15.0\text{ m s}^{-1}\). The magnitude of the resultant velocity is \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{15.0^2 + 29.7^2} = 33.3\text{ m s}^{-1}\). (c) The primary assumption made is that air resistance is negligible.

(a) [2 Marks] - Correct use of vertical kinematic equation (1 mark), Correct calculation of time to show it is approximately 3 s (1 mark). (b) [4 Marks] - Correct calculation of horizontal distance as 45.5 m (1 mark), Correct calculation of vertical component of velocity as 29.7 m s^{-1} (1 mark), Correct use of Pythagoras' theorem (1 mark), Correct final velocity magnitude of 33.3 m s^{-1} (1 mark). (c) [1.7 Marks] - States a valid assumption, such as air resistance is negligible (1.7 marks).

(a) As the mass \(M\) at the left end increases, the anticlockwise moment about rope P increases. To maintain equilibrium, the tension in rope Q must decrease. When the plank is on the verge of tipping, the tension in rope Q becomes zero, and the plank pivots about the support point of rope P at \(0.3\text{ m}\) from the left end. (b) Taking moments about rope P: the box of mass \(M\) is at a distance of \(0.3\text{ m}\) from P, and the weight of the uniform plank acts at its center of gravity, which is \(1.2\text{ m}\) from the left end (yielding a distance of \(1.2 - 0.3 = 0.9\text{ m}\) from P). At the point of tipping, the tension in Q is zero: \(M \times g \times 0.3 = 15 \times g \times 0.9\). This simplifies to \(0.3 M = 13.5\), which yields \(M = 45\text{ kg}\).

(a) [3 Marks] - States that the tension in Q decreases (1 mark), Identifies that the tension in Q becomes zero at tipping (1 mark), Identifies that the pivot is at rope P (1 mark). (b) [4.7 Marks] - Identifies the position of the center of gravity of the plank (1 mark), Calculates correct perpendicular distances to the pivot (1 mark), Sets up the correct moments equation (1 mark), Obtains the correct mass of 45 kg (1.7 marks).

(a) Since the speed is constant, the resultant force along the slope is zero. The driving force \(F\) must balance the resistive force \(R\) and the component of the car's weight acting down the slope, \(W \sin(\theta)\). Calculating weight: \(W = mg = 1200 \times 9.81 = 11772\text{ N}\). The component of weight down the slope is \(11772 \sin(8.0^\circ) = 1638\text{ N}\). Thus, the driving force is \(F = 450 + 1638 = 2088\text{ N}\) (or \(2090\text{ N}\) to 3 s.f.). (b) Power is calculated using \(P = F \times v = 2088 \times 18 = 37584\text{ W}\) (or \(37.6\text{ kW}\)).

(a) [4 Marks] - Identifies the component of weight parallel to the slope (1 mark), Correctly calculates weight component parallel to slope (1 mark), Formulates balanced force equation (1 mark), Calculates driving force as 2088 N or 2090 N (1 mark). (b) [3.7 Marks] - Recalls and uses the power formula (1 mark), Substitutes values correctly (1 mark), Correct final answer with units (1.7 marks).

(a) The tension force is \(F = mg = 4.5 \times 9.81 = 44.15\text{ N}\). Using the Young modulus formula \(E = \frac{F L}{A \Delta L}\), we rearrange for the extension \(\Delta L = \frac{F L}{A E}\). Substituting the values: \(\Delta L = \frac{44.15 \times 1.8}{2.5 \times 10^{-7} \times 1.2 \times 10^{11}} = 2.65 \times 10^{-3}\text{ m}\) (or \(2.65\text{ mm}\)). (b) Steel has a higher Young modulus than copper, which indicates that it has higher stiffness and greater resistance to elastic deformation. Since extension is inversely proportional to the Young modulus for a given force, length, and cross-sectional area, steel undergoes smaller extension.

(a) [4 Marks] - Calculates tension force correctly (1 mark), Recalls and rearranges Young modulus equation (1 mark), Substitutes correct values (1 mark), Calculates extension as 2.65 mm (1 mark). (b) [3.7 Marks] - States steel has higher Young modulus (1 mark), Explains Young modulus represents stiffness (1 mark), Explains inverse relationship between extension and Young modulus to conclude steel extends less (1.7 marks).

(a) The three forces are: weight acting downwards, upthrust acting upwards, and drag acting upwards. Terminal velocity is achieved when the resultant force on the sphere is zero, meaning the downward weight is balanced by the sum of upthrust and drag: \(W = U + F_D\). (b) The weight of the sphere is \(W = \rho_s V g\) and the upthrust is \(U = \rho_f V g\). Stoke's law gives the drag as \(F_D = 6 \pi \eta r v\). Under terminal velocity: \((\rho_s - \rho_f) V g = 6 \pi \eta r v\). Since \(V = \frac{4}{3}\pi r^3\), the formula simplifies to \(v = \frac{2 r^2 g (\rho_s - \rho_f)}{9 \eta}\). Substituting the values: \(v = \frac{2 \times (2.0 \times 10^{-3})^2 \times 9.81 \times (7800 - 920)}{9 \times 0.25} = 0.240\text{ m s}^{-1}\).

(a) [3 Marks] - Identifies the three forces and their directions (1 mark), Explains that the resultant force is zero at terminal velocity (1 mark), Formulates the correct balanced force equation (1 mark). (b) [4.7 Marks] - Uses Stoke's law and upthrust formulas (1 mark), Calculates the density difference correctly (1 mark), Rearranges equation to make velocity the subject (1 mark), Obtains the correct terminal velocity of 0.24 m s^{-1} (1.7 marks).

(a) Initial gravitational potential energy: \(E_p = mgh = 65 \times 9.81 \times 4.0 = 2550.6\text{ J}\). Final kinetic energy: \(E_k = 0.5 \times m \times v^2 = 0.5 \times 65 \times (7.5)^2 = 1828.1\text{ J}\). The work done against resistive forces is the difference in energy: \(W_{\text{resistive}} = E_p - E_k = 2550.6 - 1828.1 = 722.5\text{ J}\) (or \(723\text{ J}\)). (b) As speed increases, the air resistance opposing the motion increases. This decreases the net force acting on the skateboarder down the ramp. Since acceleration is directly proportional to net force, the acceleration decreases as speed increases.

(a) [4.7 Marks] - Calculates potential energy correctly (1 mark), Calculates kinetic energy correctly (1 mark), Identifies difference as work done (1 mark), Obtains correct final answer of 723 J (1.7 marks). (b) [3 Marks] - States air resistance increases with speed (1 mark), Explains net force down the slope decreases (1 mark), Concludes acceleration decreases as a result (1 mark).

(a) According to the conservation of momentum: \(m_A u_A + m_B u_B = (m_A + m_B) v\). Substituting the given values: \(12000 \times 3.5 = 30000 v\), which simplifies to \(42000 = 30000 v\), yielding \(v = 1.4\text{ m s}^{-1}\). (b) The initial kinetic energy is \(E_{k,\text{initial}} = 0.5 \times 12000 \times 3.5^2 = 73500\text{ J}\). The final kinetic energy is \(E_{k,\text{final}} = 0.5 \times 30000 \times 1.4^2 = 29400\text{ J}\). The loss in kinetic energy is \(73500 - 29400 = 44100\text{ J}\). Since kinetic energy is not conserved, the collision is inelastic.

(a) [3 Marks] - Recalls conservation of momentum (1 mark), Substitutes values correctly (1 mark), Calculates common speed as 1.4 m s^{-1} (1 mark). (b) [4.7 Marks] - Calculates initial kinetic energy (1 mark), Calculates final kinetic energy (1 mark), Calculates the kinetic energy loss (1 mark), Concludes that the collision is inelastic due to non-conservation of kinetic energy (1.7 marks).

(a) The loading and unloading curves form a hysteresis loop, indicating that the path followed during unloading is different from loading. For any given extension, the force during unloading is always less than during loading. The area under the loading curve represents the work done in stretching the rubber band, while the area under the unloading curve represents the work done by the rubber band. The area enclosed by the loop represents the energy dissipated as thermal energy. (b) Car suspension dampers need to absorb energy from bumps rather than returning it as kinetic energy, which would cause continuous bouncing. The hysteresis loop shows that rubber dissipates a significant portion of the work done on it as thermal energy, effectively dampening oscillations.

(a) [4.7 Marks] - Explains that the curves form a hysteresis loop (1 mark), Identifies that unloading force is smaller for a given extension (1 mark), Relates area under curves to work done (1 mark), Correctly explains that the loop area represents dissipated heat energy (1.7 marks). (b) [3 Marks] - Connects hysteresis to energy dissipation (1 mark), States that this prevents continuous bouncing (1 mark), Concludes rubber successfully dampens oscillations (1 mark).

(a) A correct diagram should have:
- A downward vertical arrow labeled 'Weight' (or \(W\), or 'force of gravity').
- Two upward vertical arrows: one labeled 'Upthrust' (or \(U\)) and one labeled 'Drag' (or 'viscous drag', or \(D\)).

(b) Initially, when the sphere is released, its velocity is zero, meaning the upward viscous drag is also zero. Since the downward weight is greater than the upward upthrust, there is a net downward resultant force, causing the sphere to accelerate downwards.
As the velocity of the sphere increases, the upward viscous drag increases. This reduces the net downward resultant force, which decreases the acceleration.
Eventually, the sum of the upward forces (Upthrust + Drag) equals the downward Weight. The net force becomes zero, acceleration becomes zero, and the sphere continues at a constant terminal velocity.

(c) At terminal velocity:
\(W = U + D\)

Substitute the formulas for weight, upthrust, and Stokes' law drag:
\(\rho_s V g = \rho_f V g + 6\pi\eta r v\)

Since \(V = \frac{4}{3}\pi r^3\):
\(\frac{4}{3}\pi r^3 g(\rho_s - \rho_f) = 6\pi\eta r v\)

Rearranging to make \(\eta\) the subject:
\(\eta = \frac{2 r^2 g (\rho_s - \rho_f)}{9 v}\)

Substitute the given values:
\(\eta = \frac{2 \times (1.50 \times 10^{-3} \text{ m})^2 \times 9.81 \text{ m s}^{-2} \times (7800 \text{ kg m}^{-3} - 950 \text{ kg m}^{-3})}{9 \times 0.112 \text{ m s}^{-1}}\)
\(\eta = \frac{2 \times (2.25 \times 10^{-6}) \times 9.81 \times 6850}{1.008}\)
\(\eta = \frac{0.30239325}{1.008} \approx 0.300 \text{ Pa s}\) (or \(\text{N s m}^{-2}\) or \(\text{kg m}^{-1} \text{s}^{-1}\))

**Part (a)**
* **MP1**: Correctly directed downward arrow labeled 'Weight' (or \(W\)) and upward arrow labeled 'Upthrust' (or \(U\)). (1)
* **MP2**: Upward arrow labeled 'Drag' / 'Viscous drag' (or \(D\) or \(F_v\)). (1)
*(Note: Arrows do not need to be scaled perfectly, but must point in correct directions. Reject if arrows do not originate from a single central point or the boundary of the sphere, or if non-physical forces like 'terminal force' are included).*

**Part (b)**
* **MP1**: Identifies that at start, drag is zero, so there is a net downward force (Weight > Upthrust) causing acceleration. (1)
* **MP2**: States that as velocity increases, viscous drag increases, which reduces the resultant force. (1)
* **MP3**: Explains that terminal velocity is reached when Upthrust + Drag = Weight, resulting in zero net force and zero acceleration. (1)

**Part (c)**
* **MP1**: Uses \(W = U + D\) to express equilibrium or states Stokes' Law equation including density terms: \(\eta = \frac{2 r^2 g (\rho_s - \rho_f)}{9 v}\). (1)
* **MP2**: Correct substitution of values into a rearranged formula. (1)
* **MP3**: Calculates \(\eta = 0.30 \text{ Pa s}\) (or \(0.300 \text{ Pa s}\)) with appropriate units. (1)
*(Accept equivalent units: \(\text{kg m}^{-1}\text{s}^{-1}\) or \(\text{N s m}^{-2}\). Allow 2 or 3 sig figs).*

The two wires are connected in series, so they must carry the same current \(I\). The relationship between current and drift velocity is given by \(I = nAvq\), where \(n\) is the number density of charge carriers, \(A\) is the cross-sectional area, \(v\) is the drift velocity, and \(q\) is the elementary charge. Since both wires are made of the same material, \(n\) is constant. The diameter of the second wire is half that of the first wire, which means its cross-sectional area \(A'\) is: \(A' = \pi \left(\frac{d}{2}\right)^2 = \frac{1}{4} \pi d^2 = \frac{A}{4}\). Since the current \(I\) is constant: \(nAvq = nA'v'q \implies Av = A'v' \implies Av = \left(\frac{A}{4}\right)v' \implies v' = 4v\).

1 mark: Correctly identifies that area decreases by a factor of 4 and drift velocity must increase by a factor of 4 to maintain constant current.

The formula for fringe spacing is \(x = \frac{\lambda D}{d}\). If the screen distance becomes \(2D\) and the slit separation becomes \(\frac{d}{2}\), the new fringe spacing \(x'\) is: \(x' = \frac{\lambda (2D)}{\left(\frac{d}{2}\right)} = 4 \left(\frac{\lambda D}{d}\right) = 4x\).

1 mark: Correct application of the fringe spacing formula showing that doubling \(D\) and halving \(d\) results in a fourfold increase in spacing.

An increase in light intensity causes the resistance of the LDR to decrease. Since the LDR is in series with a fixed resistor \(R\), a decrease in the resistance of the LDR means it takes a smaller fraction of the total supply voltage. Therefore, the output voltage \(V_{\text{out}}\) measured across the LDR decreases.

1 mark: Correctly identifies that LDR resistance decreases with intensity and that this causes a corresponding decrease in \(V_{\text{out}}\).

For a string fixed at both ends, the stationary wave patterns have nodes at both boundaries. The third harmonic corresponds to three loops (half-wavelengths) fitting within the length \(L\) of the string. Thus, \(L = 3 \times \frac{\lambda}{2} = \frac{3}{2}\lambda\). Solving for the wavelength gives \(\lambda = \frac{2}{3}L\).

1 mark: Correctly relates the length of the string to three half-wavelengths and solves for wavelength.

Einstein's photoelectric equation is \(E_{\text{k,max}} = hf - \phi\), where \(h\) is Planck's constant and \(\phi\) is the work function of the metal. Comparing this to the equation of a straight line, \(y = mx + c\), we see that plotting \(E_{\text{k,max}}\) on the y-axis and \(f\) on the x-axis gives a straight line with a gradient \(m = h\). Thus, the gradient represents Planck's constant.

1 mark: Correctly identifies the gradient of the \(E_{\text{k,max}}\) versus \(f\) graph as Planck's constant.

First, find the equivalent resistance \(R_{\text{p}}\) of the two parallel resistors: \(\frac{1}{R_{\text{p}}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \implies R_{\text{p}} = \frac{R}{2}\). Next, add the resistance of the third resistor which is connected in series: \(R_{\text{total}} = R_{\text{p}} + R = \frac{R}{2} + R = \frac{3}{2}R\).

1 mark: Correctly calculates parallel resistance as \(R/2\) and adds series resistance \(R\) to get \(\frac{3}{2}R\).

When unpolarized light of intensity \(I_0\) passes through the first polarizer, its intensity is halved, so \(I_1 = \frac{1}{2}I_0\). When this polarized light passes through the second polarizer, Malus's Law applies: \(I_2 = I_1 \cos^2\theta\). Given \(\theta = 30^\circ\), we have \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\), so \(\cos^2(30^\circ) = \frac{3}{4}\). Thus, \(I_2 = \left(\frac{1}{2}I_0\right) \left(\frac{3}{4}\right) = \frac{3}{8}I_0\).

1 mark: Correctly applies the halving of intensity for unpolarized light and Malus's law for the second polarizer.

The terminal potential difference \(V\) is equal to the e.m.f. \(E\) minus the potential drop across the internal resistance (the lost volts), which is \(Ir\). This gives the equation \(V = E - Ir\). Rearranging this equation to make \(E\) the subject yields \(E = V + Ir\).

1 mark: Correctly identifies the relation between e.m.f., terminal potential difference, and internal resistance.

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by \(E_k = hf - \phi\), which can be rearranged to express the initial photon energy: \(hf = E_k + \phi\). When the frequency of the incident light is doubled, the new maximum kinetic energy \(E_k'\) is given by: \(E_k' = h(2f) - \phi = 2(hf) - \phi\). Substituting the expression for \(hf\) into this equation yields: \(E_k' = 2(E_k + \phi) - \phi = 2E_k + \phi\). Thus, option B is the correct answer.

1 mark for the correct selection of option B. Accept explanations using the algebraic substitution of Einstein's photoelectric equation.

For a negative temperature coefficient (NTC) thermistor, a decrease in temperature results in an increase in its resistance. Since the thermistor is in series with a fixed resistor, the total resistance of the circuit increases, which causes the current in the circuit to decrease. Because the resistance of the thermistor increases while the resistance of the fixed resistor remains constant, the thermistor takes a larger share of the cell's constant electromotive force. Therefore, the potential difference across the thermistor increases. This corresponds to option C.

1 mark for the correct selection of option C. Accept reasoning linking temperature decrease to resistance increase, current decrease, and potential divider share increase.

(a) Transverse waves travel along the wire from the vibration generator and undergo reflection at the fixed end/pulley. The incident and reflected waves, which have the same frequency, wavelength, and amplitude, travel in opposite directions and superpose (interfere). Where they meet in phase, constructive interference occurs to form antinodes (regions of maximum amplitude). Where they meet in anti-phase, destructive interference occurs to form nodes (regions of zero amplitude).

(b) First, calculate the mass per unit length, \(\mu\):
\(\mu = \frac{1.24 \times 10^{-3}\text{ kg}}{0.65\text{ m}} = 1.908 \times 10^{-3}\text{ kg m}^{-1}\)

Next, calculate the speed of the transverse wave, \(v\), on the wire:
\(v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{45}{1.908 \times 10^{-3}}} \approx 153.57\text{ m s}^{-1}\)

For the first harmonic (fundamental frequency), the wavelength is related to the length \(L\) by:
\(\lambda = 2L = 2 \times 0.65\text{ m} = 1.30\text{ m}\)

Finally, calculate the frequency, \(f\):
\(f = \frac{v}{\lambda} = \frac{153.57\text{ m s}^{-1}}{1.30\text{ m}} \approx 118.1\text{ Hz}\)

Rounding to an appropriate number of significant figures gives \(118\text{ Hz}\) (or \(120\text{ Hz}\) to 2 s.f.).

(a) [Max 3 marks]
- MP1: Mention of wave reflecting at the fixed end/boundary (1)
- MP2: Mention of superposition/interference of incident and reflected waves traveling in opposite directions (1)
- MP3: State that waves have the same frequency/wavelength and amplitude, forming nodes and antinodes (1)

(b) [Max 4.7 marks]
- MP1: Correct calculation of mass per unit length \(\mu = 1.91 \times 10^{-3}\text{ kg m}^{-1}\) (1)
- MP2: Use of formula \(v = \sqrt{\frac{T}{\mu}}\) with correct substitution (1)
- MP3: State or use \(\lambda = 2L = 1.30\text{ m}\) (1)
- MP4: Use of wave equation \(v = f\lambda\) to solve for \(f\) (1)
- MP5: Correct final answer of \(118\text{ Hz}\) (allow 118 - 120 Hz) with unit (0.7)

(a) From the conservation of energy, the total electrical energy supplied per unit charge by the cell (e.m.f., \(\varepsilon\)) is equal to the sum of the potential difference across the external load resistance (terminal p.d., \(V\)) and the potential difference dropped across the internal resistance of the cell ('lost volts', \(v_{\text{lost}} = Ir\)).
Therefore, \(\varepsilon = V + Ir\)
Rearranging this gives:
\(V = \varepsilon - Ir\)

(b) Comparing the equation \(V = -rI + \varepsilon\) with the general straight-line equation \(y = mx + c\):
- The vertical intercept (y-intercept), where \(I = 0\), represents the e.m.f. \(\varepsilon\).
- The gradient of the graph represents negative internal resistance (\(-r\)), so internal resistance \(r\) is equal to the magnitude of the gradient.

(c) Set up two simultaneous equations using \(V = \varepsilon - Ir\):
1) \(1.32 = \varepsilon - 0.35r\)
2) \(1.07 = \varepsilon - 0.85r\)

Subtract equation (2) from equation (1) to eliminate \(\varepsilon\):
\(1.32 - 1.07 = -0.35r - (-0.85r)\)
\(0.25 = 0.50r\)
\(r = 0.50\ \Omega\)

(a) [Max 2 marks]
- MP1: Explain terms: total energy per unit charge (\(\varepsilon\)) equals terminal p.d. plus energy lost per unit charge across internal resistance (\(Ir\)) (1)
- MP2: Correct algebraic derivation leading to \(V = \varepsilon - Ir\) (1)

(b) [Max 2 marks]
- MP1: e.m.f. \(\varepsilon\) is equal to the y-intercept / vertical axis intercept (1)
- MP2: Internal resistance \(r\) is equal to the negative of the gradient / magnitude of the gradient (1)

(c) [Max 3.7 marks]
- MP1: Correctly substitutes both data pairs into \(V = \varepsilon - Ir\) (1)
- MP2: Eliminates \(\varepsilon\) to find an expression for \(r\) (1)
- MP3: Correct calculation of \(r = 0.50\) (1)
- MP4: Correct unit of ohms (\(\Omega\)) given (0.7)

(a) The critical angle is the angle of incidence in the more optically dense medium (glass) for which the angle of refraction in the less optically dense medium (air) is \(90^\circ\).

(b) Using the relationship between refractive index and critical angle for a boundary with air:
\(\sin \theta_c = \frac{1}{n}\)
\(\sin \theta_c = \frac{1}{1.52} \approx 0.6579\)
\(\theta_c = \sin^{-1}(0.6579) = 41.14^\circ\)
So, \(\theta_c = 41.1^\circ\)

(c) For a boundary between two media of refractive indices \(n_1\) (glass) and \(n_2\) (oil), where \(n_1 > n_2\):
\(\sin \theta_c = \frac{n_2}{n_1}\)
Substitute the given values:
\(\sin(68.0^\circ) = \frac{n_{\text{oil}}}{1.52}\)
\(n_{\text{oil}} = 1.52 \times \sin(68.0^\circ)\)
\(n_{\text{oil}} = 1.52 \times 0.9272 = 1.409\)
So, \(n_{\text{oil}} = 1.41\)

(a) [Max 2 marks]
- MP1: State that it is the angle of incidence in the denser medium (1)
- MP2: State that the corresponding angle of refraction is \(90^\circ\) (1)

(b) [Max 2 marks]
- MP1: Use of \(\sin \theta_c = \frac{1}{n}\) (1)
- MP2: Correct calculation of critical angle as \(41.1^\circ\) (or \(41^\circ\)) (1)

(c) [Max 3.7 marks]
- MP1: Use of \(\sin \theta_c = \frac{n_2}{n_1}\) (1)
- MP2: Correct substitution: \(\sin(68.0^\circ) = \frac{n_{\text{oil}}}{1.52}\) (1)
- MP3: Correct rearrangement to find \(n_{\text{oil}}\) (1)
- MP4: Correct final answer to 3 s.f. of \(1.41\) (no unit) (0.7)

(a) The photoelectric effect operates on a one-to-one interaction between a single photon and a single electron. The energy of a photon is given by \(E = \frac{hc}{\lambda}\). Increasing the wavelength decreases the photon energy. If the wavelength is greater than the threshold wavelength, the photon energy is less than the work function (the minimum energy required to release an electron). Increasing intensity only increases the rate of photons arriving but does not increase the energy of individual photons, so no individual electron can gain enough energy to escape.

(b) Convert the work function \(\Phi\) from electron-volts to Joules:
\(\Phi = 4.30\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 6.88 \times 10^{-19}\text{ J}\)

Calculate the energy \(E\) of the incident photon:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{240 \times 10^{-9}\text{ m}} = 8.2875 \times 10^{-19}\text{ J}\)

Apply Einstein's photoelectric equation to find the maximum kinetic energy, \(E_{k,\text{max}}\):
\(E_{k,\text{max}} = E - \Phi\)
\(E_{k,\text{max}} = 8.2875 \times 10^{-19}\text{ J} - 6.88 \times 10^{-19}\text{ J} = 1.4075 \times 10^{-19}\text{ J}\)
Rounding to 3 s.f. gives \(1.41 \times 10^{-19}\text{ J}\) (using standard constants, values around \(1.39 \times 10^{-19}\text{ J}\) to \(1.41 \times 10^{-19}\text{ J}\) are accepted depending on rounding intermediate steps).

(a) [Max 3.7 marks]
- MP1: State that the photon-electron interaction is a one-to-one process (1)
- MP2: Mention that energy of a photon is inversely proportional to wavelength (or \(E = hc/\lambda\)) (1)
- MP3: Explain that if wavelength is above the threshold, photon energy is less than the work function (1)
- MP4: Explain that increased intensity only increases number of photons but not individual photon energy (0.7)

(b) [Max 4 marks]
- MP1: Correct conversion of eV to Joules: \(\Phi = 6.88 \times 10^{-19}\text{ J}\) (1)
- MP2: Use of \(E = \frac{hc}{\lambda}\) to find photon energy (1)
- MP3: Correct calculation of photon energy: \(8.29 \times 10^{-19}\text{ J}\) (1)
- MP4: Apply \(E_{k,\text{max}} = E - \Phi\) to find \(1.41 \times 10^{-19}\text{ J}\) (accept range \(1.39 \times 10^{-19}\text{ J}\) to \(1.41 \times 10^{-19}\text{ J}\)) (1)

(a) As the temperature of the NTC thermistor increases, more charge carriers (conduction electrons) are released in the semiconductor material. This causes the resistance of the thermistor to decrease. In a series potential divider, a smaller resistance across the thermistor means it takes a smaller proportion of the total input voltage, resulting in a decrease in \(V_{\text{out}}\).

(b) Using the potential divider equation:
\(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{th}}}{R_{\text{th}} + R_{\text{fixed}}}\)
\(V_{\text{out}} = 9.0\text{ V} \times \frac{1.5\text{ k}\Omega}{1.5\text{ k}\Omega + 2.2\text{ k}\Omega}\)
\(V_{\text{out}} = 9.0 \times \frac{1.5}{3.7} \approx 3.65\text{ V}\)

(c) When \(V_{\text{out}} = 6.0\text{ V}\), the potential difference across the fixed resistor \(V_{\text{fixed}}\) is:
\(V_{\text{fixed}} = 9.0\text{ V} - 6.0\text{ V} = 3.0\text{ V}\)

Using the ratio of potential differences in series:
\(\frac{R_{\text{th}}}{R_{\text{fixed}}} = \frac{V_{\text{out}}}{V_{\text{fixed}}}\)
\(\frac{R_{\text{th}}}{2.2\text{ k}\Omega} = \frac{6.0\text{ V}}{3.0\text{ V}} = 2\)
\(R_{\text{th}} = 2 \times 2.2\text{ k}\Omega = 4.4\text{ k}\Omega\) (or \(4400\ \Omega\))

(a) [Max 3 marks]
- MP1: Increasing temperature increases the number density of charge carriers (1)
- MP2: Leading to a decrease in the resistance of the thermistor (1)
- MP3: Meaning the thermistor gets a smaller share of the voltage, so \(V_{\text{out}}\) decreases (1)

(b) [Max 2.7 marks]
- MP1: Correct substitution of values into potential divider formula (1)
- MP2: Correct calculation of \(V_{\text{out}} = 3.65\text{ V}\) (allow 3.6 V - 3.7 V) (1.7)

(c) [Max 2 marks]
- MP1: Calculation of the potential difference across the fixed resistor as \(3.0\text{ V}\) or setting up the ratio equation (1)
- MP2: Correct calculation of thermistor resistance as \(4.4\text{ k}\Omega\) (or \(4400\ \Omega\)) (1)

(a) Calculate the grating spacing \(d\):
\(d = \frac{1}{N} = \frac{1}{4.50 \times 10^5\text{ m}^{-1}} = 2.222 \times 10^{-6}\text{ m}\)

Apply the grating equation \(d \sin \theta = n\lambda\) for \(n = 2\):
\(\lambda = \frac{d \sin \theta}{2}\)
\(\lambda = \frac{2.222 \times 10^{-6} \times \sin(31.5^\circ)}{2}\)
\(\lambda = \frac{2.222 \times 10^{-6} \times 0.5225}{2} \approx 5.805 \times 10^{-7}\text{ m}\)
This is approximately \(5.8 \times 10^{-7}\text{ m}\).

(b) Green light has a shorter wavelength than red light (\(\lambda_{\text{green}} < \lambda_{\text{red}}\)). According to \(\sin \theta = \frac{n\lambda}{d}\), since \(\lambda\) is smaller and \(d\) remains constant, the angle of diffraction \(\theta\) for each order will be smaller. Therefore, the bright spots on the screen will be closer together.

(c) For maximum number of orders, the angle \(\theta\) cannot exceed \(90^\circ\), so \(\sin \theta \le 1\).
\(n \le \frac{d}{\lambda}\)
\(n \le \frac{2.222 \times 10^{-6}\text{ m}}{5.81 \times 10^{-7}\text{ m}} \approx 3.82\)

Since \(n\) must be an integer, the maximum order that can be observed is \(n = 3\).

(a) [Max 3 marks]
- MP1: Correctly calculate the grating spacing \(d = 2.22 \times 10^{-6}\text{ m}\) (1)
- MP2: Use of \(d \sin \theta = n\lambda\) with \(n = 2\) (1)
- MP3: Correct substitution leading to \(5.8 \times 10^{-7}\text{ m}\) (1)

(b) [Max 2 marks]
- MP1: State that green light has a shorter wavelength than red light (1)
- MP2: State that the diffraction angles are smaller, so the maxima are closer together on the screen (1)

(c) [Max 2.7 marks]
- MP1: Use of condition \(\sin \theta \le 1\) (or \(\theta \le 90^\circ\)) (1)
- MP2: Calculate maximum \(n \approx 3.82\) and state that the integer limit is 3 (1.7)

(a) First find the mean diameter \(d\):
\(d = \frac{0.45 + 0.46 + 0.44 + 0.45 + 0.45}{5} = 0.45\text{ mm} = 4.5 \times 10^{-4}\text{ m}\)

Calculate the cross-sectional area \(A\):
\(A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2\)
\(A = \pi \left(2.25 \times 10^{-4}\text{ m}\right)^2 \approx 1.59 \times 10^{-7}\text{ m}^2\)

(b) Use the resistivity formula:
\(\rho = \frac{R A}{L}\)
\(\rho = \frac{3.40\ \Omega \times 1.59 \times 10^{-7}\text{ m}^2}{1.20\text{ m}} \approx 4.505 \times 10^{-7}\ \Omega\text{ m}\)
Rounding to 2 significant figures gives \(4.5 \times 10^{-7}\ \Omega\text{ m}\).

(c) Precaution: Use a very low current / switch off the current between readings.
Explanation: This prevents the temperature of the wire from rising. Since resistivity increases with temperature, keeping the temperature constant ensures the measured resistance represents the wire at room temperature.

(a) [Max 3 marks]
- MP1: Calculate mean diameter as \(0.45\text{ mm}\) (1)
- MP2: Use of circular area formula \(A = \pi r^2\) or \(A = \frac{\pi d^2}{4}\) (1)
- MP3: Correct calculation of area as \(1.59 \times 10^{-7}\text{ m}^2\) (1)

(b) [Max 2.7 marks]
- MP1: Use of \(\rho = \frac{RA}{L}\) (1)
- MP2: Correct substitution and calculation of \(\rho = 4.5 \times 10^{-7}\ \Omega\text{ m}\) with correct unit (1.7)

(c) [Max 2 marks]
- MP1: Correct precaution stated (e.g., keep current low, switch off between readings, avoid stretching the wire) (1)
- MP2: Valid explanation matching the precaution (e.g., temperature changes change resistance, stretching decreases cross-sectional area) (1)

(a) In unpolarized light, the oscillations of the electric field vector occur in all possible planes perpendicular to the direction of propagation. In polarized light, the oscillations are restricted to a single plane that contains the direction of propagation.

(b) Place one polarizing filter in front of the light source to produce plane-polarized light. Then, place a second polarizing filter (the analyzer) behind the first. Rotate the second filter through \(360^\circ\). Observe that the intensity of the transmitted light varies, dropping to zero (complete extinction) when the transmission axes of the two filters are perpendicular (at \(90^\circ\)). This demonstrates that light waves are transverse, as only transverse waves can be polarized.

(c) When unpolarized light of intensity \(I_0\) passes through the first polarizing filter, its intensity is halved because only one component of the oscillations is transmitted:
\(I_1 = 0.5 I_0\)

When this polarized light passes through the second filter, Malus's Law applies:
\(I_2 = I_1 \cos^2 \theta\)
\(I_2 = (0.5 I_0) \cos^2(35.0^\circ)\)
Calculate the cosine term:
\(\cos(35.0^\circ) \approx 0.81915\)
\(\cos^2(35.0^\circ) \approx 0.6710\)

Now, calculate the final intensity:
\(I_2 = 0.5 I_0 \times 0.6710 \approx 0.3355 I_0\)
Thus, the intensity is approximately \(0.336 I_0\) (or \(0.34 I_0\) to 2 s.f.).

(a) [Max 2 marks]
- MP1: State that unpolarized light has oscillations in all planes perpendicular to propagation (1)
- MP2: State that polarized light has oscillations in a single plane perpendicular to propagation (1)

(b) [Max 3 marks]
- MP1: Describe placing two filters in series (polarizer and analyzer) (1)
- MP2: Describe rotating one filter relative to the other (1)
- MP3: State that observation of variation in intensity / extinction at \(90^\circ\) proves the wave is transverse (1)

(c) [Max 2.7 marks]
- MP1: State or show that after the first filter, intensity is half: \(I_1 = 0.5 I_0\) (1)
- MP2: Apply Malus's law: \(I = I_1 \cos^2 \theta\) (1)
- MP3: Correct calculation to give \(0.336 I_0\) (or \(0.34 I_0\)) (0.7)

(a) The work function is defined as the minimum energy required to liberate an electron from the surface of a metal.

(b) First, convert the work function from electron-volts (eV) to Joules (J):
\( \Phi = 3.68 \text{ eV} \times 1.60 \times 10^{-19} \text{ J/eV} = 5.888 \times 10^{-19} \text{ J} \)

Next, calculate the energy of the incident photons:
\( E = hf = 6.63 \times 10^{-34} \text{ J s} \times 1.25 \times 10^{15} \text{ Hz} = 8.2875 \times 10^{-19} \text{ J} \)

Apply Einstein's photoelectric equation:
\( hf = \Phi + E_{k\text{,max}} \)
\( E_{k\text{,max}} = hf - \Phi \)
\( E_{k\text{,max}} = 8.2875 \times 10^{-19} \text{ J} - 5.888 \times 10^{-19} \text{ J} = 2.40 \times 10^{-19} \text{ J} \)

(c)(i) There is no change to the maximum kinetic energy. The maximum kinetic energy depends only on the frequency of the radiation (and work function), which remains constant.

(c)(ii) The rate of emission of photoelectrons increases. Increasing intensity means more photons hit the metal surface per second. Since there is a 1-to-1 interaction between photons and surface electrons, more electrons are emitted per second.

**Part (a): [2 Marks]**
* **MP1:** Minimum energy required to remove/emits/release an electron [1 Mark]
* **MP2:** From the surface of a metal / from the metal [1 Mark]
*(Do not accept "to move an electron through the metal")*

**Part (b): [3 Marks]**
* **MP1:** Use of \( W = eV \) to convert work function to Joules (\( 5.89 \times 10^{-19} \text{ J} \)) [1 Mark]
* **MP2:** Use of \( E = hf \) to find photon energy (\( 8.29 \times 10^{-19} \text{ J} \)) [1 Mark]
* **MP3:** Correct subtraction to find \( E_{k\text{,max}} = 2.40 \times 10^{-19} \text{ J} \) (accept \( 2.4 \times 10^{-19} \text{ J} \)) [1 Mark]

**Part (c)(i): [1 Mark]**
* **MP1:** No change / remains the same because the energy of individual photons (or frequency) is unchanged [1 Mark]

**Part (c)(ii): [2 Marks]**
* **MP1:** Rate of emission increases / more electrons are emitted per second [1 Mark]
* **MP2:** Because higher intensity means more photons incident per second, and there is a 1-to-1 interaction between photons and electrons [1 Mark]

(a) Independent variable: Distance \( s \).
Dependent variable: Time \( t \).

(b) Repeat the measurement of \( t \) multiple times (at least three times) for each distance \( s \) and calculate a mean. Ensure the sphere is released from the exact same position above the first light gate each time to keep the initial speed zero.

(c) For \( s = 0.500 \text{ m} \):
\( t^2 = (0.320 \text{ s})^2 = 0.102 \text{ s}^2 \)

For \( s = 1.000 \text{ m} \):
\( t^2 = (0.452 \text{ s})^2 = 0.204 \text{ s}^2 \)

(d) Since the sphere is released from rest, the initial velocity \( u = 0 \). The equation of motion simplifies to:
\( s = \frac{1}{2}gt^2 \)
Comparing this to the equation of a straight line \( y = mx + c \) where \( y = s \) and \( x = t^2 \):
- The y-intercept \( c = 0 \), so the line passes through the origin.
- The gradient \( m = \frac{1}{2}g \).
Therefore, the acceleration of free fall is \( g = 2 \times \text{gradient} \).

(e) Given \( \text{gradient} = 4.85 \pm 0.12 \text{ m s}^{-2} \):
\( g = 2 \times 4.85 = 9.70 \text{ m s}^{-2} \).
Since \( g = 2m \), the percentage uncertainty in \( g \) is equal to the percentage uncertainty in the gradient:
Percentage uncertainty = \( \frac{0.12}{4.85} \times 100\\% = 2.47\\% \approx 2.5\\% \).

Part (a):
- Correctly identifies independent variable as distance and dependent variable as time (or \( t^2 \)). [1.5 marks]

Part (b):
- Mentions repeating measurements and calculating a mean. [1 mark]
- Mentions ensuring consistent release position or using an electromagnet. [1 mark]

Part (c):
- Correct calculation of first \( t^2 \) value to 3 s.f. (0.102). [1 mark]
- Correct calculation of second \( t^2 \) value to 3 s.f. (0.204). [1 mark]
- Correct units (\( \text{s}^2 \)). [0.5 marks]

Part (d):
- States that \( u = 0 \) so the intercept is zero. [1 mark]
- Relates gradient of \( s \) vs \( t^2 \) to \( g/2 \). [1 mark]
- Concludes that \( g = 2 \times \text{gradient} \). [0.5 marks]

Part (e):
- Correct calculation of \( g = 9.70 \text{ m s}^{-2} \). [2 marks]
- Correct calculation of percentage uncertainty as \( 2.5\\% \) (accept \( 2.47\\% \)). [2 marks]

(a) A long, thin wire will undergo a much larger, more easily measurable extension \( x \) for any given tension. This significantly reduces the percentage uncertainty in the extension measurements.

(b) Safety precautions:
1. Wear safety goggles to protect eyes in case the wire snaps under high tension.
2. Place a padded box or heavy tray on the floor below the suspended masses to safely catch them if the wire breaks.

(c) Young Modulus is defined as:
\( E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{x / L} = \frac{F L}{A x} \)
Since the force is due to gravity on the mass, \( F = mg \).
Since the wire is cylindrical, the cross-sectional area is \( A = \frac{\pi d^2}{4} \).
Substituting these gives:
\( E = \frac{mgL}{\left(\frac{\pi d^2}{4}\right)x} = \frac{4gL}{\pi d^2} \left(\frac{m}{x}\right) \)
Since the gradient of a plot of \( m \) against \( x \) is \( \frac{m}{x} \), we have:
\( E = \frac{4gL}{\pi d^2} \times \text{gradient} \).

(d) Convert diameter to meters: \( d = 0.35 \times 10^{-3} \text{ m} \).
Using \( g = 9.81 \text{ m s}^{-2} \):
\( E = \frac{4 \times 9.81 \times 2.80}{\pi \times (0.35 \times 10^{-3})^2} \times 385 \)
\( E = \frac{109.872}{3.8485 \times 10^{-7}} \times 385 \)
\( E = 2.855 \times 10^8 \times 385 = 1.10 \times 10^{11} \text{ Pa} \).

(e) Percentage uncertainty in diameter \( d \):
\( \frac{0.01}{0.35} \times 100\\% = 2.86\\% \).
Since \( A = \frac{\pi d^2}{4} \), the area is proportional to \( d^2 \).
Therefore, the percentage uncertainty in area \( A \) is:
\( 2 \times 2.86\\% = 5.71\\% \approx 5.7\\% \).

Part (a):
- Identifies that extension is larger for a long/thin wire under the same force. [1 mark]
- Explains that a larger extension leads to a smaller percentage uncertainty. [1.5 marks]

Part (b):
- Mentions safety goggles. [1 mark]
- Mentions box/padding to catch falling masses. [1 mark]

Part (c):
- Correctly defines Stress, Strain, and Young Modulus. [1 mark]
- Correctly substitutes formula for area \( A = \pi d^2 / 4 \) and force \( F = mg \). [1 mark]
- Correctly links the term \( m/x \) to the gradient of the graph to show the requested formula. [1 mark]

Part (d):
- Correct conversion of diameter to meters and substitution of values. [1 mark]
- Correct intermediate steps/use of \( g = 9.81 \text{ m s}^{-2} \). [1 mark]
- Correct final value \( 1.10 \times 10^{11} \text{ Pa} \) (accept range \( 1.09 \times 10^{11} \text{ to } 1.11 \times 10^{11} \text{ Pa} \)). [1 mark]

Part (e):
- Correctly calculates percentage uncertainty in diameter (\( 2.86\\% \)). [1 mark]
- Correctly doubles this value to find the percentage uncertainty in area (\( 5.7\\% \) or \( 5.71\\% \)). [1 mark]

(a) The test wire should be connected in a series circuit containing a variable DC power supply, an ammeter (to measure current), and a switch. A voltmeter must be connected in parallel across the specific length \( l \) of the wire being measured. This allows resistance \( R \) to be calculated using \( R = V/I \).

(b) A switch is included to prevent current flowing through the wire when readings are not being taken. Continuous current causes the wire to heat up, which would increase its temperature and consequently increase its resistance, leading to systematic errors.

(c) Calculating \( R/l \) for each length:
- \( R_1/l_1 = 1.18 / 0.250 = 4.72 \Omega \text{ m}^{-1} \)
- \( R_2/l_2 = 2.34 / 0.500 = 4.68 \Omega \text{ m}^{-1} \)
- \( R_3/l_3 = 3.56 / 0.750 = 4.75 \Omega \text{ m}^{-1} \)
- \( R_4/l_4 = 4.70 / 1.000 = 4.70 \Omega \text{ m}^{-1} \)

Mean value of \( R/l \):
\( \text{Mean} = \frac{4.72 + 4.68 + 4.75 + 4.70}{4} = 4.71 \Omega \text{ m}^{-1} \).

(d) Cross-sectional area:
\( A = \frac{\pi d^2}{4} = \frac{\pi \times (0.36 \times 10^{-3} \text{ m})^2}{4} = 1.02 \times 10^{-7} \text{ m}^2 \) (or \( 1.018 \times 10^{-7} \text{ m}^2 \)).
Percentage uncertainty in \( d \):
\( \frac{0.02}{0.36} \times 100\\% = 5.56\\% \).
Percentage uncertainty in \( A \):
\( 2 \times 5.56\\% = 11.1\\% \).

(e) Resistivity is given by:
\( \rho = R \frac{A}{l} = \left(\frac{R}{l}\right) A = 4.71 \times 1.018 \times 10^{-7} = 4.79 \times 10^{-7} \Omega \text{ m} \) (or \( 4.8 \times 10^{-7} \Omega \text{ m} \)).
Absolute uncertainty in \( \rho \) is:
\( 11.1\\% \times 4.79 \times 10^{-7} = 0.53 \times 10^{-7} \Omega \text{ m} \) (or \( 0.5 \times 10^{-7} \Omega \text{ m} \)).
Thus, \( \rho = (4.8 \pm 0.5) \times 10^{-7} \Omega \text{ m} \).

Part (a):
- Ammeter connected in series with the power supply and wire. [1 mark]
- Voltmeter connected in parallel across the measured length of the wire. [1 mark]
- Switch included in the main circuit loop. [1 mark]

Part (b):
- Identifies that continuous current heats the wire. [1 mark]
- Explains that increased temperature increases the resistance (and resistivity), affecting accuracy. [1 mark]

Part (c):
- Correctly calculates all four ratios of \( R/l \). [1 mark]
- Calculates the correct mean of \( 4.71 \Omega \text{ m}^{-1} \). [1 mark]
- Includes the correct unit (\( \Omega \text{ m}^{-1} \) or Ohm per meter). [1 mark]

Part (d):
- Calculates the area correctly as \( 1.02 \times 10^{-7} \text{ m}^2 \). [1 mark]
- Calculates the percentage uncertainty in diameter as \( 5.56\\% \). [0.5 marks]
- Doubles this value to find the percentage uncertainty in area as \( 11.1\\% \) (or \( 11\\% \)). [1 mark]

Part (e):
- Correct value of resistivity \( 4.8 \times 10^{-7} \Omega \text{ m} \). [1 mark]
- Correctly calculates the absolute uncertainty as \( 0.5 \times 10^{-7} \Omega \text{ m} \) (or \( 0.53 \times 10^{-7} \text{ m} \)). [1 mark]

(a) The laser is aligned to shine directly through the diffraction grating onto a flat vertical screen. To ensure normal incidence, the student can check that the reflected ray from the grating goes directly back into the laser aperture, or ensure that the positions of the diffracted order spots on the screen are completely symmetrical on both sides of the central maximum.

(b) Do not look directly into the laser beam, and avoid pointing the laser at shiny/reflective surfaces to prevent accidental specular reflections into the eyes.

(c) Measuring a larger distance (double the value) using the same instrument (with a fixed absolute uncertainty) halves the percentage uncertainty of the measurement. It also avoids the difficulty of identifying the exact center of the bright central spot as the single starting reference point.

(d) The mean distance from the central maximum to a first-order maximum is:
\( y = \frac{91.6 \text{ cm}}{2} = 45.8 \text{ cm} = 0.458 \text{ m} \).
Using trigonometry for the right-angled triangle formed by the grating, central maximum, and first-order maximum:
\( \tan \theta = \frac{y}{D} = \frac{0.458 \text{ m}}{2.40 \text{ m}} = 0.1908 \)
\( \theta = \arctan(0.1908) = 10.80^\circ \).

(e) The grating has \( 300 \text{ lines per mm} \), which is \( 300,000 \text{ lines per meter} \).
Slit spacing:
\( d = \frac{1}{300,000 \text{ lines m}^{-1}} = 3.333 \times 10^{-6} \text{ m} \).

Using the grating equation for \( n = 1 \):
\( d \sin \theta = n \lambda \)
\( \lambda = d \sin \theta = (3.333 \times 10^{-6} \text{ m}) \times \sin(10.80^\circ) \)
\( \lambda = 3.333 \times 10^{-6} \times 0.1874 = 6.25 \times 10^{-7} \text{ m} = 625 \text{ nm} \).

Part (a):
- Identifies alignment of laser, grating, and screen. [1 mark]
- Explains how to ensure normal incidence (e.g., checking back reflection or symmetry of diffracted spots). [2 marks]

Part (b):
- Mentions not looking directly at the beam or avoiding reflections. [1 mark]

Part (c):
- Explains that measuring a larger distance with the same absolute uncertainty reduces percentage uncertainty. [1 mark]
- Mentions that it reduces the error in identifying the center of the central spot. [1 mark]

Part (d):
- Calculates \( y = 0.458 \text{ m} \). [0.5 marks]
- Uses \( \tan \theta = y/D \) correctly. [1 mark]
- Arrives at \( \theta = 10.8^\circ \) (or \( 10.80^\circ \)). [1 mark]

Part (e):
- Calculates \( d = 3.33 \times 10^{-6} \text{ m} \). [1.5 marks]
- Substitutes \( d \), \( \theta \), and \( n = 1 \) into \( d \sin \theta = n \lambda \). [1 mark]
- Arrives at \( \lambda = 625 \text{ nm} \) (accept range \( 620 \text{ nm to } 630 \text{ nm} \) due to rounding). [1.5 marks]