2025 CCEA AS-Level Physics 1210 模擬試題連答案詳解

A scalar quantity has magnitude only (size), whereas a vector quantity has both magnitude and direction. Example of a scalar used in mechanics: speed (or mass, energy, distance). Example of a vector used in mechanics: velocity (or displacement, force, acceleration, momentum).

[1] correct distinction between scalar and vector (magnitude only vs magnitude and direction); [1] one correct example of each named. Accept any valid mechanics scalar/vector pair; reject 'distance' as a vector or 'displacement' as a scalar.

Newton's third law states that when body A exerts a force on body B, body B exerts an equal and opposite force on body A, and the two forces act on different objects. When the swimmer's hand pushes the water backwards, the water simultaneously pushes the swimmer's hand forwards with a force of equal magnitude. Since this reaction force acts on the swimmer (not the water), it is this unbalanced forward force on the swimmer that produces a forward acceleration.

[1] correct statement of Newton's third law (equal and opposite forces on different bodies); [1] correct application: reaction force from water acts forward on swimmer, causing forward acceleration. Reject answers that only restate 'equal and opposite' without identifying the forward force acting on the swimmer.

The EMF, \( \varepsilon \), is the total electrical energy converted from chemical (or other) energy per unit charge driven around the complete circuit, including the energy dissipated inside the cell itself. The terminal p.d., \( V \), is the energy transferred to the external circuit per unit charge only. When current \( I \) flows through a cell of internal resistance \( r \), \( V = \varepsilon - Ir \), so the terminal p.d. is always less than the EMF whenever current flows (equal only when \( I = 0 \)).

[1] EMF correctly defined as total energy per unit charge (including energy dissipated internally); [1] terminal p.d. correctly defined as energy per unit charge delivered to the external circuit, with reference to \( V = \varepsilon - Ir \) or equivalent statement that it is less than EMF when current flows.

A potential divider is used to produce a variable or fixed fraction of a supply voltage (a chosen output voltage smaller than the supply). For two resistors \( R_1 \) and \( R_2 \) in series across a supply \( V_{in} \), the same current flows through both, so the voltage across each is proportional to its resistance: \( V_{out} = V_{in}\left(\dfrac{R_2}{R_1+R_2}\right) \). The output voltage therefore depends only on the ratio \( R_2 : (R_1+R_2) \), not on the absolute resistance values.

[1] correct statement of purpose (produces a fraction/chosen value of the supply voltage); [1] correct relationship \( V_{out}=V_{in}R_2/(R_1+R_2) \) or equivalent statement that output voltage is proportional to the fraction of total resistance across which it is measured.

(a) Vertically, the stone falls freely from rest: \( s = \tfrac{1}{2}gt^2 \), so \( t = \sqrt{\dfrac{2s}{g}} = \sqrt{\dfrac{2(45.0)}{9.81}} = 3.03\text{ s} \).
(b) Horizontal motion is unaffected by gravity, so \( x = v_x t = 15.0 \times 3.03 = 45.4\text{ m} \).
(c) The vertical velocity at landing is \( v_y = gt = 9.81 \times 3.03 = 29.7\text{ m s}^{-1} \). The resultant speed is \( v = \sqrt{v_x^2+v_y^2} = \sqrt{15.0^2+29.7^2} = \sqrt{225+882.1} = 33.3\text{ m s}^{-1} \).
(d) \( \theta = \tan^{-1}\left(\dfrac{v_y}{v_x}\right) = \tan^{-1}\left(\dfrac{29.7}{15.0}\right) = 63.2^{\circ} \) below the horizontal.
Final answer: \( v = 33.3\text{ m s}^{-1} \) at \( 63.2^{\circ} \) below the horizontal.

(a) [1] correct rearrangement \( t=\sqrt{2s/g} \); [1] \( t = 3.03\text{ s} \) (accept 3.0 s). (b) [1] use of \( x=v_xt \) with their t; [1] \( x = 45.4\text{ m} \) (ECF from (a)). (c) [1] \( v_y=gt \) evaluated (29.7 m/s); [1] correct use of Pythagoras; [1] \( v = 33.3\text{ m s}^{-1} \) (ECF). (d) [1] correct use of \( \tan\theta = v_y/v_x \); [1] \( \theta = 63.2^{\circ} \) (ECF). Arithmetic slip penalised once only, on the final numerical answer.

(a) By conservation of momentum: \( m_1u_1 = (m_1+m_2)v \), so \( 2.4 \times 3.5 = (2.4+1.6)v \), giving \( v = \dfrac{8.4}{4.0} = 2.1\text{ m s}^{-1} \).
(b) \( KE_i = \tfrac{1}{2}(2.4)(3.5)^2 = 14.7\text{ J} \). \( KE_f = \tfrac{1}{2}(4.0)(2.1)^2 = 8.82\text{ J} \). Loss \( = 14.7 - 8.82 = 5.88\text{ J} \).
(c) The impulse on the \( 1.6\text{ kg} \) trolley equals its change in momentum: \( \Delta p = 1.6 \times (2.1-0) = 3.36\text{ N s} \); by Newton's third law this equals the magnitude of the impulse the \( 2.4\text{ kg} \) trolley exerts on it.
(d) The collision is inelastic, since kinetic energy is lost (not conserved), even though momentum is conserved.

(a) [1] correct conservation of momentum equation; [1] correct substitution; [1] \( v = 2.1\text{ m s}^{-1} \). (b) [1] correct \( KE_i = 14.7\text{ J} \); [1] correct \( KE_f = 8.82\text{ J} \) (ECF from (a)); [1] loss \( = 5.88\text{ J} \). (c) [1] correct use of \( \Delta p = m_2 v \); [1] \( 3.36\text{ N s} \) (ECF). (d) [1] 'inelastic' with correct reason (KE not conserved / KE is lost). Reject 'elastic' even if momentum reasoning is correct.

(a) The weight of a uniform beam acts at its midpoint, i.e. \( 2.0\text{ m} \) from A (and from B).
(b) The weight of the beam acts \( 2.0 - 1.2 = 0.8\text{ m} \) from the pivot (on the B side), and the load \( W \) acts \( 1.2\text{ m} \) from the pivot (on the A side). Taking moments about the pivot for equilibrium: sum of clockwise moments = sum of anticlockwise moments, so \( W \times 1.2 = 150 \times 0.8 \), giving \( W = \dfrac{120}{1.2} = 100\text{ N} \).
(c) Since the beam is in vertical equilibrium, the upward reaction at the pivot equals the total downward force: \( R = W + 150 = 100 + 150 = 250\text{ N} \).

(a) [1] midpoint / centre of the beam (2.0 m from each end). (b) [1] correct perpendicular distance for the beam's weight (0.8 m); [1] correct perpendicular distance for W (1.2 m); [1] correct moments equation \( W \times 1.2 = 150 \times 0.8 \); [1] correct rearrangement; [1] \( W = 100\text{ N} \). (c) [1] recognition that \( R = W + 150 \) (vertical equilibrium); [1] correct substitution (ECF); [1] \( R = 250\text{ N} \).

Weight \( W = mg = 25 \times 9.81 = 245\text{ N} \).
(a) Component down the slope \( = W\sin20^{\circ} = 245 \times \sin20^{\circ} = 83.9\text{ N} \).
(b) Normal reaction \( N = W\cos20^{\circ} = 245 \times \cos20^{\circ} = 230\text{ N} \).
(c) Friction \( F = \mu N = 0.30 \times 230 = 69.1\text{ N} \), acting down the slope (opposing the crate's motion up the slope).
(d) At constant velocity the crate is in equilibrium along the slope, so \( T = W\sin20^{\circ} + F = 83.9 + 69.1 = 153\text{ N} \).

(a) [1] use of \( W\sin20^{\circ} \); [1] \( 83.9\text{ N} \) (accept 83.6-84.1 N for g=9.8-9.81). (b) [1] use of \( W\cos20^{\circ} \); [1] \( 230\text{ N} \). (c) [1] use of \( F=\mu N \) (ECF from (b)); [1] \( 69.1\text{ N} \). (d) [1] recognition that constant velocity means zero resultant force along the slope; [1] correct equation \( T=W\sin20^{\circ}+F \); [1] \( T=153\text{ N} \) (ECF).

(a) Radius \( = 0.23\text{ mm} = 2.3\times10^{-4}\text{ m} \). \( A = \pi r^2 = \pi(2.3\times10^{-4})^2 = 1.66\times10^{-7}\text{ m}^2 \).
(b) \( R = \dfrac{\rho L}{A} = \dfrac{4.9\times10^{-7}\times2.50}{1.66\times10^{-7}} = 7.37\ \Omega \).
(c) \( V = IR = 0.85 \times 7.37 = 6.27\text{ V} \).
(d) At constant volume, doubling the length halves the cross-sectional area (since \( V = AL \) is constant). New resistance \( R' = \dfrac{\rho(2L)}{(A/2)} = 4 \times \dfrac{\rho L}{A} = 4R = 4 \times 7.37 = 29.5\ \Omega \).

(a) [1] correct radius used (halving diameter); [1] \( A=1.66\times10^{-7}\text{ m}^2 \). (b) [1] correct formula \( R=\rho L/A \); [1] correct substitution (ECF); [1] \( R=7.37\ \Omega \). (c) [1] use of \( V=IR \) (ECF); [1] \( V=6.27\text{ V} \). (d) [1] correct reasoning that area halves as length doubles at constant volume; [1] \( R'=4R=29.5\ \Omega \) (ECF). Accept direct recomputation via \( \rho(2L)/(A/2) \) for full credit.

(a) \( R_{total} = 15+25 = 40\ \Omega \).
(b) \( I = \dfrac{V}{R} = \dfrac{12.0}{40} = 0.30\text{ A} \).
(c) \( V_{25\Omega} = IR = 0.30\times25 = 7.5\text{ V} \).
(d) \( t = 5.0\times60 = 300\text{ s} \); \( Q = It = 0.30\times300 = 90\text{ C} \).
(e) \( P = VI = 12.0\times0.30 = 3.6\text{ W} \) (check: \( I^2R = 0.30^2\times40 = 3.6\text{ W} \), consistent).

(a) [1] \( 40\ \Omega \). (b) [1] use of \( I=V/R \) (ECF); [1] \( I=0.30\text{ A} \). (c) [1] use of \( V=IR \) with 25 Ω (ECF); [1] \( V=7.5\text{ V} \). (d) [1] correct conversion of time to seconds; [1] \( Q=90\text{ C} \) (ECF). (e) [1] correct power formula used (\( P=VI \) or \( I^2R \)); [1] \( P=3.6\text{ W} \) (ECF). Unit penalty of at most [1] applies once across the whole question for a missing/incorrect unit.

(a) \( \Delta PE = mgh = 65\times9.81\times22 = 1.40\times10^4\text{ J} \).
(b) By conservation of energy, \( \tfrac{1}{2}mv^2 = \Delta PE \), so \( v = \sqrt{\dfrac{2\Delta PE}{m}} = \sqrt{2\times9.81\times22} = 20.8\text{ m s}^{-1} \) (note the mass cancels, giving the same result as \( v=\sqrt{2gh} \), which confirms the calculation).
(c) In practice there is friction between the skis and snow, and air resistance acting on the skier; both do negative work on the skier, converting some of the kinetic energy into heat/sound, so the actual kinetic energy (and hence speed) at the bottom is lower than the frictionless-model value.

(a) [1] correct formula \( mgh \); [1] correct substitution; [1] \( \Delta PE=1.40\times10^4\text{ J} \). (b) [1] correct energy conservation equation \( \tfrac12mv^2=\Delta PE \); [1] correct rearrangement (ECF); [1] \( v=20.8\text{ m s}^{-1} \). (c) [1] identifies friction and/or air resistance as present in reality; [1] correct consequence explained (energy dissipated as heat, so less KE / lower speed). Reject an answer that says the skier speeds up more in reality.

Taking the ball's initial direction of travel as positive: initial momentum \( p_i = 0.058\times22 = 1.276\text{ kg m s}^{-1} \); final momentum \( p_f = 0.058\times(-18) = -1.044\text{ kg m s}^{-1} \) (since the ball rebounds, its velocity reverses sign).
(a) \( \Delta p = p_f - p_i = -1.044 - 1.276 = -2.32\text{ kg m s}^{-1} \); magnitude \( 2.32\text{ kg m s}^{-1} \) (i.e. \( 2.32\text{ N s} \)).
(b) \( F = \dfrac{\Delta p}{\Delta t} = \dfrac{2.32}{5.5\times10^{-3}} = 422\text{ N} \).
(c) By Newton's third law, the force the ball exerts on the wall is equal in magnitude and opposite in direction to the force the wall exerts on the ball: \( 422\text{ N} \), directed horizontally into the wall.

(a) [1] correct momentum before (1.276 kg m/s); [1] correct momentum after, with sign reversal (-1.044 kg m/s); [1] \( \Delta p=2.32\text{ kg m s}^{-1} \). (b) [1] correct conversion of 5.5 ms to seconds; [1] use of \( F=\Delta p/\Delta t \) (ECF); [1] \( F=422\text{ N} \). (c) [1] correct magnitude (ECF, equal to (b)); [1] correct direction (into the wall) with reference to Newton's third law.

(a) The table shows R increasing in direct proportion to l (e.g. doubling l from 0.20 m to 0.40 m doubles R from 0.51 Ω to 1.02 Ω). A graph of R against l would therefore be a straight line passing through the origin, showing \( R \propto l \).
(b) Using two well-separated points, e.g. \( (0.20, 0.51) \) and \( (1.00, 2.55) \): gradient \( = \dfrac{2.55-0.51}{1.00-0.20} = \dfrac{2.04}{0.80} = 2.55\ \Omega\text{ m}^{-1} \).
(c) Since \( R = \dfrac{\rho l}{A} \), the gradient of R against l equals \( \dfrac{\rho}{A} \). So \( \rho = \text{gradient}\times A = 2.55\times8.0\times10^{-8} = 2.0\times10^{-7}\ \Omega\text{ m} \).
(d) The student should use as long a length of wire as practical, and measure it precisely (e.g. using a set square against a metre rule, or a travelling microscope) to reduce the percentage uncertainty in the length measurement.

(a) [1] straight line through the origin identified/implied; [1] correctly links this to direct proportionality between R and l. (b) [1] correct method using two points on the line (or the ratio R/l); [1] \( 2.55\ \Omega\text{ m}^{-1} \) (accept 2.5-2.6). (c) [1] correct relation between gradient and \( \rho/A \); [1] correct substitution (ECF); [1] \( \rho=2.0\times10^{-7}\ \Omega\text{ m} \) (accept 1.9-2.1 ×10⁻⁷). (d) [1] any valid precaution addressing measurement precision/uncertainty in length. Reject vague answers such as 'be careful'.

(a) \( v^2 \) values: \( 1.0^2=1.0,\ 2.0^2=4.0,\ 3.0^2=9.0,\ 4.0^2=16.0,\ 5.0^2=25.0 \) (all in \( \text{m}^2\text{s}^{-2} \)).
(b) Since \( KE = \tfrac{1}{2}mv^2 \), plotting KE against \( v^2 \) gives a straight line through the origin with gradient \( \tfrac{1}{2}m \). Checking the data, \( KE/v^2 \) is constant at \( 0.75\text{ J s}^2\text{m}^{-2} \) for every row (e.g. \( 18.75/25.0 = 0.75 \)), so the gradient is \( 0.75 \), representing half the mass of the trolley, \( \tfrac{1}{2}m \).
(c) \( m = 2\times\text{gradient} = 2\times0.75 = 1.5\text{ kg} \), which matches the mass of the trolley stated in the question (1.5 kg), confirming the relationship \( KE=\tfrac12mv^2 \) and the validity of the gradient method.
(d) A graph of KE against v would be a curve (parabola), from which it is difficult to measure a gradient accurately or to test the relationship precisely. Plotting KE against \( v^2 \) linearises the relationship into a straight line through the origin, whose constant gradient can be measured accurately and used to determine \( m \) directly.

(a) [1] all five values correct (1.0, 4.0, 9.0, 16.0, 25.0). (b) [1] correct gradient value 0.75 (from at least one calculation shown); [1] correctly identifies gradient as constant/consistent across the data; [1] correctly states gradient represents \( \tfrac12m \). (c) [1] correct method \( m=2\times\text{gradient} \); [1] \( m=1.5\text{ kg} \) with correct comparison to given mass. (d) [1] identifies that KE vs v is a curve / non-linear; [1] identifies that KE vs v² is linear, allowing accurate gradient determination.

(a) \( \dfrac{1}{R_{23}} = \dfrac{1}{20}+\dfrac{1}{30} \Rightarrow R_{23} = \dfrac{20\times30}{20+30} = \dfrac{600}{50} = 12\ \Omega \).
(b) \( R_{total} = R_1 + R_{23} + r = 10+12+0.50 = 22.5\ \Omega \).
(c) \( I = \dfrac{\varepsilon}{R_{total}} = \dfrac{9.00}{22.5} = 0.400\text{ A} \).
(d) Terminal p.d. \( V = \varepsilon - Ir = 9.00 - (0.400\times0.50) = 9.00-0.20 = 8.80\text{ V} \) (check: this should equal \( I\times(R_1+R_{23}) = 0.400\times22 = 8.80\text{ V} \), which it does — confirms consistency).
(e) Voltage across the parallel section \( V_{23} = I\times R_{23} = 0.400\times12 = 4.80\text{ V} \). Current through \( R_2 \): \( I_2 = \dfrac{V_{23}}{R_2} = \dfrac{4.80}{20} = 0.24\text{ A} \). (Check: current through \( R_3 \) is \( 4.80/30=0.16\text{ A} \); \( 0.24+0.16=0.40\text{ A} \), matching the total current from (c), confirming the split is correct.)
(f) \( P_{ext} = VI = 8.80\times0.400 = 3.52\text{ W} \) (check: \( I^2(R_1+R_{23}) = 0.400^2\times22 = 3.52\text{ W} \), consistent).
(g) Total time \( t = 2.0\times3600 = 7200\text{ s} \). Total energy from the EMF: \( E = \varepsilon I t = 9.00\times0.400\times7200 = 2.592\times10^4\text{ J} \). Energy dissipated in the external resistors: \( E_{ext}=P_{ext}t = 3.52\times7200 = 2.534\times10^4\text{ J} \). Energy dissipated in the internal resistance: \( E_{int}=I^2rt = 0.400^2\times0.50\times7200 = 576\text{ J} \). Sum: \( 2.534\times10^4+576 = 2.592\times10^4\text{ J} \), which equals the total EMF energy calculated above, confirming conservation of energy.

(a) [1] correct parallel formula; [1] \( R_{23}=12\ \Omega \). (b) [1] correct summation including r; [1] \( 22.5\ \Omega \) (ECF). (c) [1] use of \( I=\varepsilon/R_{total} \); [1] \( I=0.400\text{ A} \) (ECF). (d) [1] correct use of \( V=\varepsilon-Ir \); [1] \( V=8.80\text{ V} \) (ECF). (e) [1] correct calculation of \( V_{23} \) or equivalent method; [1] \( I_2=0.24\text{ A} \) (ECF). (f) [1] correct method (\( P=VI \) or \( I^2R \) for external circuit); [1] \( P=3.52\text{ W} \) (ECF). (g) [1] correct total EMF energy \( 2.59\times10^4\text{ J} \); [1] correct external energy \( 2.53\times10^4\text{ J} \); [1] correct internal energy 576 J; [1] correct demonstration that the two sums are equal (energy conservation), with a valid concluding statement. ECF applied throughout from earlier parts.

(a) The wavelength is the distance between two adjacent points on a wave that are in phase with each other, e.g. the distance between successive crests (or successive troughs).
(b) The frequency is the number of complete oscillations (or wavelengths) passing a fixed point per unit time (per second).
(c) The wave equation is \( v = f\lambda \).

(a) [1] correct definition referencing adjacent points in phase (e.g. crest to crest). (b) [1] correct definition referencing oscillations/wavelengths per second. (c) [1] correct equation \( v=f\lambda \) stated. Accept equivalent correct wording throughout.

(a) The principle of superposition states that when two or more waves meet at a point, the resultant displacement at that point is the vector sum (algebraic sum, for waves along the same line) of the displacements that each wave would have produced separately.
(b) Constructive interference occurs at P when the path difference between the two waves arriving at P is a whole number of wavelengths: path difference \( = n\lambda \), where \( n = 0, 1, 2, 3, \ldots \)

(a) [1] correct statement of the principle of superposition (resultant = vector/algebraic sum of individual displacements). (b) [1] path difference expressed in terms of \( \lambda \); [1] correct condition \( n\lambda \) with n as a whole number (integer, including zero). Reject 'path difference = 0' as the sole answer (too restrictive).

(a) The refractive index of a medium is defined as \( n = \dfrac{c}{c_{medium}} \), the ratio of the speed of light in a vacuum to the speed of light in the medium.
(b) Snell's law: \( n_1\sin\theta_1 = n_2\sin\theta_2 \), where \( n_1 \) and \( n_2 \) are the refractive indices of medium 1 and medium 2, and \( \theta_1 \), \( \theta_2 \) are the angles between the ray and the normal to the boundary in medium 1 and medium 2 respectively.

(a) [1] correct definition \( n=c/c_{medium} \) (ratio of speeds, correct way up). (b) [1] correct equation \( n_1\sin\theta_1=n_2\sin\theta_2 \); [1] all symbols correctly defined (angles measured from the normal). Accept \( \sin\theta_1/\sin\theta_2 = n_2/n_1 \) as an equivalent form.

(a) The photoelectric effect is the emission of electrons from the surface of a metal (or other material) when electromagnetic radiation (e.g. ultraviolet or visible light) of sufficiently high frequency is incident upon it.
(b) The threshold frequency \( f_0 \) is the minimum frequency of incident radiation required to cause photoemission from a given metal. For incident radiation with frequency below \( f_0 \), no photoelectrons are emitted no matter how great the intensity of the radiation, because each individual photon does not carry enough energy to overcome the work function of the metal.

(a) [1] correct reference to emission of electrons from a metal surface; [1] correct condition (incident EM radiation of sufficiently high frequency). (b) [1] correctly identifies \( f_0 \) as the minimum frequency for emission to occur; [1] correctly explains that below \( f_0 \) no electrons are emitted regardless of intensity (one-photon-one-electron reasoning).

Apparatus: A laser is set up on a bench so that its beam is incident normally on a diffraction grating with a known number of lines per metre, N (so the line spacing is \( d = 1/N \)). A screen is placed a measured distance D from the grating, parallel to the grating, in a darkened room so the diffraction pattern of bright spots can be seen clearly.

Measurements: The laser is switched on, producing a pattern of bright spots (orders) on the screen either side of a central (zero-order) maximum. For each visible order n (n = 1, 2, 3, …), the distance \( x_n \) from the central maximum to the nth-order spot is measured with a metre rule, and the perpendicular distance D from the grating to the screen is also measured.

Theory/calculation: The angle of diffraction for the nth order is found from \( \tan\theta_n = \dfrac{x_n}{D} \). This angle is substituted into the diffraction grating equation \( n\lambda = d\sin\theta_n \), which is rearranged to give \( \lambda = \dfrac{d\sin\theta_n}{n} \).

Improving accuracy/reliability: Measure \( x_n \) for as many orders as are visible and calculate \( \lambda \) for each order, then take a mean value of \( \lambda \), which reduces the effect of random error in any one measurement. Use as large a value of D as practical, so that the spot separations \( x_n \) are large compared with the uncertainty in a single length measurement, reducing the percentage uncertainty. Ensure the grating is mounted perpendicular to the incident beam (checked with a set square) so that the zero order is not displaced. Repeat each distance measurement and average.

Apparatus and setup [3]: [1] laser and grating (with known line spacing/lines per metre) correctly described; [1] screen placed a measured distance D from grating; [1] darkened room / grating perpendicular to beam noted. Measurement procedure [4]: [1] identifies measurement of \( x_n \), the distance from central to nth-order maximum; [1] identifies measurement of D; [1] correctly links these via \( \tan\theta_n = x_n/D \); [1] states that this is done for more than one order n. Theory and calculation [3]: [1] correctly states the diffraction grating equation \( n\lambda=d\sin\theta \); [1] correctly rearranges for \( \lambda \); [1] correctly notes \( d=1/N \) from the given lines-per-metre value. Improving accuracy/reliability [3]: any three of: repeat/average over several orders or repeats [1]; use larger D to increase x and reduce percentage uncertainty [1]; ensure grating perpendicular to beam using a set square [1]; use vernier scale/travelling microscope or careful metre-rule technique for x [1] (max 3).

(a) From the top of the object, one ray is drawn parallel to the principal axis; after passing through the lens it is refracted so that it passes through the focal point on the far side of the lens. A second ray is drawn from the top of the object straight through the centre of the lens, which passes through undeviated. The point where these two refracted rays cross (on the far side of the lens, since the object is beyond f) marks the position of the top of the image; the image lies on the principal axis level with this point.
(b) With \( f=10\text{ cm} \), \( u=15\text{ cm} \): \( \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{3-2}{30} = \dfrac{1}{30} \), so \( v = 30\text{ cm} \).
(c) Magnification \( m = \dfrac{v}{u} = \dfrac{30}{15} = 2.0 \). Image height \( = m \times \text{object height} = 2.0\times3.0 = 6.0\text{ cm} \).
(d) Since v is positive (image on the opposite side of the lens from the object) the image is real; since \( m>1 \) it is magnified; a real image formed by a single converging lens with the object beyond f is inverted.

(a) [1] correctly describes the ray parallel to the axis refracting through the focal point; [1] correctly describes the ray through the centre being undeviated; [1] correctly states the image is located where the two refracted rays intersect. (b) [1] correct lens formula quoted; [1] correct substitution of f and u; [1] correct combination of fractions; [1] \( v=30\text{ cm} \). (c) [1] correct magnification formula \( m=v/u \); [1] \( m=2.0 \) (ECF); [1] image height \( =6.0\text{ cm} \) (ECF). (d) [1] each for real, inverted, magnified (any three valid, correctly justified where a reason is asked) — [1] mark reserved for correct overall consistency/reasoning linking sign of v and value of m to the three characteristics.

(a) At the critical angle, the refracted ray grazes the boundary at \( 90^{\circ} \): \( n\sin\theta_c = 1\times\sin90^{\circ} \), so \( \sin\theta_c = \dfrac{1}{n} = \dfrac{1}{1.52} = 0.658 \), giving \( \theta_c = 41.1^{\circ} \).
(b) Since \( 38^{\circ} < \theta_c \), the ray refracts (partially) into the air. Using Snell's law from glass to air: \( n\sin38^{\circ} = 1\times\sin\theta_2 \), so \( \sin\theta_2 = 1.52\times\sin38^{\circ} = 1.52\times0.6157 = 0.936 \), giving \( \theta_2 = 69.4^{\circ} \).
(c) Since \( 45^{\circ} > \theta_c = 41.1^{\circ} \), no light can refract out into the air; instead all of the light is reflected back into the glass, with the angle of reflection equal to the angle of incidence (\( 45^{\circ} \)). This phenomenon is called total internal reflection.
(d) The core of an optical fibre has a higher refractive index than the surrounding cladding. Light is launched into the fibre so that it always strikes the core-cladding boundary at an angle of incidence greater than the critical angle for that boundary. Every time the light ray reaches the boundary it therefore undergoes total internal reflection rather than refracting out into the cladding, so (in the ideal case) no light energy escapes through the sides of the fibre; the light is repeatedly reflected along the length of the core and emerges only at the far end, transmitting the signal with minimal loss.

(a) [1] correct condition (refraction angle = 90°) or correct formula \( \sin\theta_c=1/n \); [1] correct substitution; [1] \( \theta_c=41.1^{\circ} \). (b) [1] recognises ray refracts (angle < critical angle); [1] correct use of Snell's law glass-to-air; [1] \( \theta_2=69.4^{\circ} \). (c) [1] recognises \( 45^{\circ}>\theta_c \); [1] states all light is reflected, none transmitted; [1] correctly states angle of reflection = angle of incidence (45°); [1] names 'total internal reflection'. (d) [1] core has higher refractive index than cladding; [1] light strikes core-cladding boundary at greater than critical angle; [1] correctly links this to repeated total internal reflection; [1] correctly concludes light stays confined to the core / travels the fibre length with minimal loss.

(a) The work function is the minimum photon energy needed to just release an electron, corresponding to the threshold frequency: \( \phi = hf_0 = 6.63\times10^{-34}\times4.40\times10^{14} = 2.92\times10^{-19}\text{ J} \). In eV: \( \phi = \dfrac{2.92\times10^{-19}}{1.60\times10^{-19}} = 1.82\text{ eV} \).
(b) \( f = \dfrac{c}{\lambda} = \dfrac{3.00\times10^8}{400\times10^{-9}} = 7.50\times10^{14}\text{ Hz} \).
(c) Photon energy: \( E = hf = 6.63\times10^{-34}\times7.50\times10^{14} = 4.97\times10^{-19}\text{ J} \). By Einstein's photoelectric equation, \( KE_{max} = E - \phi = 4.97\times10^{-19} - 2.92\times10^{-19} = 2.06\times10^{-19}\text{ J} \).
(d) At the stopping potential, all the kinetic energy of the most energetic electrons is converted to electrical potential energy: \( eV_s = KE_{max} \), so \( V_s = \dfrac{KE_{max}}{e} = \dfrac{2.06\times10^{-19}}{1.60\times10^{-19}} = 1.28\text{ V} \).
(e) The maximum kinetic energy would be unchanged. Increasing the intensity of monochromatic light increases the number of photons arriving per second, and so increases the rate of photoelectron emission (photocurrent), but does not change the energy of an individual photon, which depends only on its frequency (\( E=hf \)); since \( KE_{max}=hf-\phi \) depends only on frequency and work function, it is unaffected by intensity.

(a) [1] correct use of \( \phi=hf_0 \); [1] \( \phi=2.92\times10^{-19}\text{ J} \); [1] correct conversion to \( 1.82\text{ eV} \). (b) [1] correct use of \( f=c/\lambda \); [1] \( f=7.50\times10^{14}\text{ Hz} \). (c) [1] correct photon energy \( E=hf=4.97\times10^{-19}\text{ J} \) (ECF); [1] correct use of \( KE_{max}=E-\phi \); [1] \( KE_{max}=2.06\times10^{-19}\text{ J} \) (ECF). (d) [1] correct relation \( eV_s=KE_{max} \); [1] correct rearrangement (ECF); [1] \( V_s=1.28\text{ V} \). (e) [1] correctly states \( KE_{max} \) is unchanged; [1] correctly explains photon energy depends on frequency only, not intensity; [1] correctly links increased intensity to more photons per second / greater photocurrent, not greater photon energy.

(a) The work done accelerating the electron equals its gain in kinetic energy: \( KE = eV = 1.60\times10^{-19}\times2.50\times10^3 = 4.00\times10^{-16}\text{ J} \).
(b) \( KE=\tfrac12 m_ev^2 \Rightarrow v = \sqrt{\dfrac{2KE}{m_e}} = \sqrt{\dfrac{2\times4.00\times10^{-16}}{9.11\times10^{-31}}} = 2.96\times10^7\text{ m s}^{-1} \) (about 10% of c, so the non-relativistic assumption is reasonable).
(c) Momentum \( p = m_ev = 9.11\times10^{-31}\times2.96\times10^7 = 2.70\times10^{-23}\text{ kg m s}^{-1} \). De Broglie wavelength: \( \lambda = \dfrac{h}{p} = \dfrac{6.63\times10^{-34}}{2.70\times10^{-23}} = 2.46\times10^{-11}\text{ m} \).
(d) The de Broglie wavelength is the wavelength \( \lambda=h/p \) associated with the matter wave of any moving particle of momentum p; it describes the wave-like behaviour that particles such as electrons can exhibit. Evidence supporting the wave nature of electrons includes electron diffraction: a beam of electrons passed through a thin polycrystalline graphite film (or the Davisson-Germer experiment using a nickel crystal) produces a diffraction pattern of concentric rings/fringes, which can only be explained if the electrons behave as waves.
(e) Since \( KE=eV \) and \( v=\sqrt{2KE/m_e} \), \( v \propto \sqrt{V} \); since \( \lambda = h/(m_ev) \), \( \lambda \propto 1/v \propto 1/\sqrt{V} \). Doubling V therefore multiplies \( \lambda \) by a factor of \( 1/\sqrt{2} \approx 0.707 \), decreasing it to about \( 2.46\times10^{-11}\times0.707 = 1.74\times10^{-11}\text{ m} \). So increasing the accelerating voltage decreases the de Broglie wavelength.

(a) [1] correct formula \( KE=eV \); [1] \( KE=4.00\times10^{-16}\text{ J} \). (b) [1] correct rearrangement of \( KE=\tfrac12mv^2 \); [1] correct substitution (ECF); [1] \( v=2.96\times10^7\text{ m s}^{-1} \). (c) [1] correct momentum \( p=m_ev \) (ECF); [1] correct use of \( \lambda=h/p \); [1] \( \lambda=2.46\times10^{-11}\text{ m} \) (ECF). (d) [1] correct definition \( \lambda=h/p \) referencing matter waves; [1] correctly names electron diffraction (or Davisson-Germer) as evidence; [1] correctly links this evidence to wave-like interference/diffraction behaviour. (e) [1] correctly identifies \( \lambda \propto 1/\sqrt{V} \); [1] correctly calculates the factor \( 1/\sqrt2 \) or new wavelength (ECF); [1] correct concluding statement that \( \lambda \) decreases.

(a) \( \Delta\lambda = 492.7-486.1 = 6.6\text{ nm} \). \( z = \dfrac{\Delta\lambda}{\lambda} = \dfrac{6.6}{486.1} = 0.0136 \).
(b) For \( v \ll c \), \( z \approx \dfrac{v}{c} \), so \( v = zc = 0.0136\times3.00\times10^8 = 4.07\times10^6\text{ m s}^{-1} \). (The increase in wavelength shows the galaxy is moving away from us — a redshift.)
(c) By Hubble's law, \( v = H_0d \), so \( d = \dfrac{v}{H_0} = \dfrac{4.07\times10^6}{2.4\times10^{-18}} = 1.70\times10^{24}\text{ m} \).

(a) [1] correct \( \Delta\lambda=6.6\text{ nm} \); [1] \( z=0.0136 \) (accept 0.0135-0.0136). (b) [1] correct use of \( v=zc \); [1] correct substitution (ECF); [1] \( v=4.07\times10^6\text{ m s}^{-1} \), with recognition this indicates recession (redshift). (c) [1] correct use of Hubble's law \( v=H_0d \); [1] correct rearrangement \( d=v/H_0 \) (ECF); [1] \( d=1.70\times10^{24}\text{ m} \).

(a) Using \( \dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v} \): for \( u=20.0,\ v=20.1 \): \( f = 10.0\text{ cm} \); for \( u=25.0,\ v=16.7 \): \( f = 10.0\text{ cm} \); for \( u=30.0,\ v=15.0 \): \( f = 10.0\text{ cm} \) (all three pairs give a consistent value, which supports the reliability of the data).
(b) Mean \( f = \dfrac{10.0+10.0+10.0}{3} = 10.0\text{ cm} \).
(c) As the observer moves their eye from side to side while looking at the search pin against the image, if the pin is not exactly at the position of the image, the pin will appear to move relative to the image (parallax). The pin is adjusted back and forth until, viewed from different eye positions, the pin and the image no longer show any relative movement — at this point the pin coincides with the true image position, and v can be read off.
(d) Both u and v should be measured from the optical centre of the lens (marked on the lens holder, or found from the midpoint of the lens holder if it is thin), not from an edge of the mount, otherwise a constant offset (systematic/index error) is introduced into every reading of u and v.

(a) [1] correct method shown for at least one pair; [1] two further correct values (allow ECF from a small arithmetic slip carried once); [1] all three values consistent at 10.0 cm (to 3 s.f.). (b) [1] correct mean calculated from their three values; [1] mean quoted as \( 10.0\text{ cm} \). (c) [1] correctly describes moving the eye from side to side; [1] correctly describes pin appearing to move relative to image before parallax is removed; [1] correctly states no relative movement indicates parallax removed / pin coincides with image. (d) [1] identifies measuring from the optical centre of the lens as the precaution; [1] correctly explains this avoids a constant/systematic offset (index error) in u and v.

(a) \( T = \dfrac{t}{20} \): for \( L=0.400\text{ m} \), \( T=\dfrac{25.3}{20}=1.27\text{ s} \); for \( L=0.600\text{ m} \), \( T=\dfrac{31.0}{20}=1.55\text{ s} \); for \( L=0.800\text{ m} \), \( T=\dfrac{35.8}{20}=1.79\text{ s} \).
(b) \( T^2 \): \( 1.27^2=1.60\text{ s}^2 \); \( 1.55^2=2.40\text{ s}^2 \); \( 1.79^2=3.20\text{ s}^2 \).
(c) Rearranging \( T^2=\dfrac{4\pi^2}{g}L \) gives \( g=\dfrac{4\pi^2L}{T^2} \). Using \( L=0.800\text{ m} \), \( T^2=3.20\text{ s}^2 \): \( g = \dfrac{4\pi^2\times0.800}{3.20} = 9.86\text{ m s}^{-2} \), which is close to the accepted value of \( 9.81\text{ m s}^{-2} \), confirming the method is sound.
(d) The student should time a large number of oscillations (e.g. 20, as done here) rather than a single oscillation, and divide the total time by that number to find T. The absolute uncertainty in starting and stopping the stopwatch (due to reaction time) is roughly the same regardless of how many oscillations are timed, so timing many oscillations makes this fixed absolute uncertainty a much smaller percentage of the (now much larger) total time measured, reducing the percentage uncertainty in T.

(a) [1] correct method \( T=t/20 \) applied; [1] all three T values correct (1.27, 1.55, 1.79 s). (b) [1] correct squaring method; [1] all three \( T^2 \) values correct (ECF from (a)). (c) [1] correct rearrangement \( g=4\pi^2L/T^2 \); [1] correct substitution using L=0.800 m data (ECF); [1] \( g=9.86\text{ m s}^{-2} \) (accept 9.8-9.9 m/s²). (d) [1] identifies timing many oscillations (not one) as the precaution; [1] correctly explains the reaction-time/absolute error is roughly fixed; [1] correctly explains this reduces the percentage uncertainty in T because the total time measured is larger.

(a) Since the micrometer reads \( +0.02\text{ mm} \) with nothing between the jaws, this positive zero error must be subtracted from every raw reading: corrected diameter \( = 12.04 - 0.02 = 12.02\text{ mm} \).
(b) Radius \( r = \dfrac{12.02}{2} = 6.01\text{ mm} = 6.01\times10^{-3}\text{ m} \); length \( l = 50.20\text{ mm} = 50.20\times10^{-3}\text{ m} \). Volume of a cylinder: \( V=\pi r^2 l = \pi\times(6.01\times10^{-3})^2\times(50.20\times10^{-3}) = 5.70\times10^{-6}\text{ m}^3 \).
(c) Mass \( m = 44.90\text{ g} = 44.90\times10^{-3}\text{ kg} \). Density \( \rho = \dfrac{m}{V} = \dfrac{44.90\times10^{-3}}{5.70\times10^{-6}} = 7880\text{ kg m}^{-3} \) (this is close to the density of steel, a physically plausible value for a metal cylinder).
(d) A real cylinder may not be perfectly uniform in cross-section along its length (e.g. very slightly tapered or out-of-round). Taking several diameter readings at different points and different orientations and averaging reduces the effect of random measurement error and gives a mean diameter more representative of the cylinder as a whole, improving the accuracy of the volume (and hence density) calculated.

(a) [1] correctly identifies that the zero error should be subtracted (since it is positive); [1] corrected diameter \( =12.02\text{ mm} \). (b) [1] correct use of radius = half of corrected diameter, converted to metres (ECF); [1] correct formula \( V=\pi r^2 l \) applied with l converted to metres; [1] \( V=5.70\times10^{-6}\text{ m}^3 \) (accept 5.6-5.8 ×10⁻⁶). (c) [1] correct conversion of mass to kg; [1] correct use of \( \rho=m/V \) (ECF); [1] \( \rho=7880\text{ kg m}^{-3} \) (accept 7700-8000, ECF). (d) [1] identifies possible non-uniformity/irregularity in the cylinder; [1] correctly explains averaging gives a more representative/accurate mean value, reducing random error.

(a) \( R_{app} = \dfrac{V}{I} = \dfrac{4.20}{0.220} = 19.1\ \Omega \).
(b) The ammeter, being external to the voltmeter-R parallel combination, measures the total current leaving the supply, which is the sum of the current through R and the (small but non-zero) current taken by the voltmeter. The voltmeter reads the true p.d. across R accurately, but the ammeter reading is slightly larger than the true current through R alone. Since \( R_{app}=V/I_{measured} \) uses a current that is too large, \( R_{app} \) is slightly smaller than the true resistance of R.
(c) When R is small, the current through R, \( I_R=V/R \), is relatively large, while the current taken by the voltmeter, \( I_V=V/R_V \), stays the same (small) regardless of R. The extra current \( I_V \) is therefore a small fraction of the total measured current when R is small, so the systematic error introduced is proportionally small. In the alternative arrangement, the ammeter (which has a small but non-zero resistance) would be placed directly in series with R, so the voltmeter would read the p.d. across R and the ammeter; for a small R, the ammeter's resistance would not be negligible compared with R, and this would introduce a proportionally much larger systematic error in V. The arrangement given is therefore the better choice for a small resistance.
(d) The student should repeat the measurement at several different values of current (by varying the supply voltage/using a rheostat), plot a graph of V against I, and take the gradient of the best-fit line as the value of R; this averages out random errors in individual readings and gives a more precise value than a single pair of readings.

(a) [1] correct use of \( R=V/I \); [1] \( R_{app}=19.1\ \Omega \). (b) [1] correctly identifies that the ammeter reads current through R plus current through voltmeter; [1] correctly identifies the voltmeter reads the true p.d. across R; [1] correctly concludes \( R_{app} \) is an underestimate as a result. (c) [1] correctly explains that voltmeter current is proportionally small compared with a large current through a small R; [1] correctly explains that in the alternative arrangement the ammeter's resistance would not be negligible compared with a small R; [1] correct overall conclusion that the given arrangement introduces the smaller systematic error for small R. (d) [1] any valid suggestion to reduce random error (e.g. repeat readings at several currents / plot V-I graph and use gradient / use digital meters); [1] correct explanation of why this improves precision.

(a) The data show P increasing from \( R=1.0\ \Omega \) to a peak somewhere between \( R=2.0\ \Omega \) and \( R=3.0\ \Omega \) (where P is almost equal, 3.56 W and 3.57 W), then decreasing steadily as R increases further to \( 8.0\ \Omega \). The curve therefore rises to a single, rounded maximum and then falls — it is not a straight line.
(b) Since P is almost identical at \( R=2.0\ \Omega \) (3.56 W) and \( R=3.0\ \Omega \) (3.57 W), the maximum of the smooth curve through the points lies close to the midpoint of this interval, at \( R \approx 2.5\ \Omega \), with a maximum power of about \( P_{max}\approx3.6\text{ W} \).
(c) Since theory predicts \( P \) is a maximum when \( R=r \), the internal resistance is estimated as \( r \approx 2.5\ \Omega \) (the value of R found in (b)).
(d) Using \( r=2.5\ \Omega \) and \( \varepsilon = 6.00\text{ V} \) at \( R=2.0\ \Omega \): \( P = \dfrac{6.00^2\times2.0}{(2.0+2.5)^2} = \dfrac{72.0}{20.25} = 3.56\text{ W} \). This agrees almost exactly with the measured value of \( 3.56\text{ W} \) at \( R=2.0\ \Omega \) in the table, confirming that \( r=2.5\ \Omega \) is a good estimate of the battery's internal resistance.

(a) [1] correctly identifies the curve rises then falls (has a maximum), not linear; [1] correctly locates the region of the maximum (between about 2.0 and 3.0 Ω). (b) [1] correctly identifies P is very similar at R=2.0 and 3.0 Ω; [1] correct estimate \( R\approx2.5\ \Omega \) (accept 2.0-3.0 Ω range with justification); [1] \( P_{max}\approx3.6\text{ W} \) (ECF). (c) [1] correctly quotes/uses the condition \( R=r \) for maximum power; [1] \( r\approx2.5\ \Omega \) (ECF from (b)). (d) [1] correct substitution into \( P=\varepsilon^2R/(R+r)^2 \) (ECF for r); [1] correct evaluation of \( (R+r)^2=20.25 \); [1] \( P=3.56\text{ W} \); [1] valid comparison made with the measured 3.56 W, noting close agreement supports the value of r found.

(a) Dividing each cm value by 100: \( x = 0.021,\ 0.040,\ 0.062,\ 0.081,\ 0.100\text{ m} \).
(b) Since \( F=kx \), a graph of F against x is a straight line through the origin with gradient k, the spring constant.
(c) Using the first point \( (0.021,1.0) \) and last point \( (0.100,5.0) \): gradient \( = \dfrac{5.0-1.0}{0.100-0.021} = \dfrac{4.0}{0.079} = 50.6\text{ N m}^{-1} \). So \( k \approx 50.6\text{ N m}^{-1} \).
(d) Elastic potential energy stored: \( E = \tfrac{1}{2}kx^2 = \tfrac{1}{2}\times50.6\times(0.100)^2 = 0.253\text{ J} \).
(e) The student should use a larger range of loads, giving larger extensions; the absolute uncertainty in reading the ruler (e.g. ±0.1 cm) then represents a smaller percentage of each (larger) extension measured, and the wider spread of x values on the graph also allows the gradient to be determined more precisely.

(a) [1] all five values correctly converted to metres. (b) [1] correctly identifies the gradient as the spring constant k. (c) [1] correct use of the two chosen points; [1] correct subtraction of x values (0.079 m); [1] correct division; [1] \( k\approx50.6\text{ N m}^{-1} \) (accept 49-52 N/m, or a valid value from a different two-point/regression method with correct working). (d) [1] correct formula \( E=\tfrac12kx^2 \) used (ECF); [1] \( E=0.253\text{ J} \) (accept 0.24-0.26 J, ECF). (e) [1] identifies using larger loads/extensions (or a longer/less stiff spring) as the modification; [1] correctly explains this reduces the percentage uncertainty in x (and hence in the gradient).

(a) Percentage uncertainty in d \( = \dfrac{0.01}{0.46}\times100 = 2.17\% \).
(b) Percentage uncertainty in L \( = \dfrac{0.01}{2.50}\times100 = 0.40\% \).
(c) Percentage uncertainty in R \( = \dfrac{0.15}{7.37}\times100 = 2.04\% \).
(d) Since \( \rho = \dfrac{\pi d^2R}{4L} \) is a product/quotient of d² , R and L (with \( \pi \) and 4 exact), the percentage uncertainties in each are added, with the percentage uncertainty in d doubled because it is squared: \( \%\Delta\rho = 2(2.17\%) + 2.04\% + 0.40\% = 4.35\%+2.04\%+0.40\% = 6.78\% \, (\approx6.8\%) \).
(e) \( \rho = \dfrac{\pi(0.46\times10^{-3})^2(7.37)}{4(2.50)} = 4.90\times10^{-7}\ \Omega\text{ m} \). Absolute uncertainty \( = \rho\times\dfrac{\%\Delta\rho}{100} = 4.90\times10^{-7}\times0.0678 = 0.33\times10^{-7}\ \Omega\text{ m} \). So \( \rho = (4.9\pm0.3)\times10^{-7}\ \Omega\text{ m} \) (quoted to 2 significant figures, matching the precision implied by the uncertainty).
(f) The diameter d contributes most. Its own percentage uncertainty (2.17%) is similar in size to that of R (2.04%), but because d is squared in the formula for \( \rho \), its contribution to the total is doubled to 4.35%, which is nearly two-thirds of the total 6.78% percentage uncertainty in \( \rho \) — more than either R or L contribute alone.
(g) The most effective improvement would be to take several diameter measurements at different points along the wire's length and in different orientations (e.g. rotating the micrometer 90° at each point) and use the mean diameter. Since d makes the largest (and doubled) contribution to the overall uncertainty, reducing the uncertainty in d — for example by averaging out random variation and any slight non-uniformity in the wire's cross-section — has the greatest effect on reducing the overall percentage uncertainty in \( \rho \), more so than making a similar improvement to the measurement of L or R.

(a) [1] correct method \( 0.01/0.46 \); [1] \( 2.17\% \). (b) [1] correct method; [1] \( 0.40\% \). (c) [1] correct method; [1] \( 2.04\% \). (d) [1] correctly identifies that percentage uncertainties are added for a product/quotient; [1] correctly doubles the percentage uncertainty in d (since it is squared); [1] correct summation shown; [1] \( 6.78\% \) (accept 6.7-6.8%, ECF from (a)-(c)). (e) [1] correct calculation of \( \rho=4.90\times10^{-7}\ \Omega\text{m} \) (or equivalent, e.g. 4.9×10⁻⁷); [1] correct method for absolute uncertainty (their %Δρ applied to their ρ); [1] \( 0.3\times10^{-7}\ \Omega\text{m} \) (ECF); [1] correctly expressed in the required \( (\ldots\pm\ldots)\times10^{-7} \) form; [1] appropriate number of significant figures (2 s.f. on ρ, consistent with the uncertainty). (f) [1] correctly identifies d as the dominant contributor; [1] correctly explains this is because d is squared (doubling its %unc); [1] correct numerical comparison/justification (e.g. 4.35% vs 2.04% vs 0.40%). (g) [1] valid, specific procedural change targeting d; [1] correct reasoning that d dominates and is doubled, so improving it has the greatest effect; [1] additional valid detail (e.g. averaging over several points/orientations); [1] correct overall coherent justification linking the change to a reduced %Δρ. Accept an equally well-justified alternative answer that correctly identifies and targets the dominant uncertainty.

(a) Using \( (0.20, 5.61) \) and \( (1.00, 4.02) \): gradient \( = \dfrac{4.02-5.61}{1.00-0.20} = \dfrac{-1.59}{0.80} = -1.99\text{ V A}^{-1} \) (≈ -2.00 V/A). Using \( V=mI+c \) with the first point: \( c = 5.61-(-1.99\times0.20) = 5.61+0.40 = 6.01\text{ V} \).
(b) Comparing with \( V=\varepsilon-Ir \), the intercept (value of V when I=0) is the EMF \( \varepsilon \), and the magnitude of the gradient is the internal resistance r. So \( \varepsilon \approx 6.01\text{ V} \) and \( r \approx 2.00\ \Omega \) (or 1.99 Ω).
(c) Percentage difference \( = \dfrac{|6.01-6.10|}{6.10}\times100 = \dfrac{0.09}{6.10}\times100 = 1.5\% \).

(a) [1] correct gradient calculation (≈ -1.99 to -2.00 V/A); [1] correct method for intercept using their gradient and a data point; [1] intercept \( \approx6.01\text{ V} \) (accept 5.9-6.1 V, ECF). (b) [1] correctly identifies intercept = EMF and gradient magnitude = internal resistance; [1] correct values quoted (ECF from (a)). (c) [1] correct use of the percentage difference formula relative to the stated 6.10 V; [1] \( \approx1.5\% \) (ECF, accept 1.3-1.7%).