2024 CCEA A-Level Physics 1210 模擬試題連答案詳解

(a) Stress is the force applied per unit cross-sectional area, \( \sigma = \frac{F}{A} \), unit Pa. Strain is the extension per unit original length, \( \varepsilon = \frac{x}{L} \), no unit.

(b) \( A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.42\times10^{-3}}{2}\right)^2 = 1.39\times10^{-7}\ \text{m}^2 \)

(c) \( \sigma = \frac{F}{A} = \frac{40.0}{1.39\times10^{-7}} = 2.89\times10^{8}\ \text{Pa} \)

(d) \( \varepsilon = \frac{x}{L} = \frac{4.04\times10^{-3}}{2.800} = 1.44\times10^{-3} \)

(e) \( E = \frac{\sigma}{\varepsilon} = \frac{2.89\times10^{8}}{1.44\times10^{-3}} = 2.00\times10^{11}\ \text{Pa} \)

(f) \( E_{strain} = \frac{1}{2}Fx = \frac{1}{2}(40.0)(4.04\times10^{-3}) = 8.08\times10^{-2}\ \text{J} \)

(g) The wire obeys Hooke's law up to this extension, i.e. the force is directly proportional to extension (the elastic limit has not been exceeded), so the load-extension graph is a straight line through the origin.

Final answer: \( E = 2.00\times10^{11}\ \text{Pa} \) (200 GPa).

(a) [1] stress = force per unit (cross-sectional) area; [1] strain = extension per unit original length. (b) [1] correct substitution into \( A=\pi(d/2)^2 \); [1] \( A = 1.39\times10^{-7}\ \text{m}^2 \) (accept 1.38–1.40 x10^-7). (c) [1] correct substitution \( \sigma=F/A \); [1] \( \sigma = 2.89\times10^{8}\ \text{Pa} \) (ECF from (b)). (d) [1] correct substitution \( \varepsilon=x/L \); [1] \( \varepsilon = 1.44\times10^{-3} \). (e) [1] \( E=\sigma/\varepsilon \); [1] \( E = 2.00\times10^{11}\ \text{Pa} \), accept 1.95–2.05 x10^11 (ECF from (c) and (d)); reject answer with no unit or wrong power of ten. (f) [1] correct substitution into \( \frac12 Fx \); [1] \( 8.08\times10^{-2}\ \text{J} \) (accept 80–81 mJ). (g) [1] states Hooke's law / proportionality holds (linear load–extension) up to this point; reject vague 'elastic' with no reference to proportionality.

(a) For a fixed mass of gas, \( \frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \), so \( p_2 = p_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 1.01\times10^{5} \times \frac{2.40\times10^{-3}}{1.80\times10^{-3}} \times \frac{320}{290} = 1.49\times10^{5}\ \text{Pa} \)

(b) \( pV = nRT \Rightarrow n = \frac{p_1V_1}{RT_1} = \frac{(1.01\times10^{5})(2.40\times10^{-3})}{(8.31)(290)} = 0.101\ \text{mol} \)

(c) \( N = nN_A = 0.101 \times 6.02\times10^{23} = 6.06\times10^{22} \) molecules

(d) \( \frac{3}{2}kT = \frac{3}{2}(1.38\times10^{-23})(290) = 6.00\times10^{-21}\ \text{J} \)

(e) The mean kinetic energy of a molecule is directly proportional to the absolute (kelvin) temperature \( \left(\overline{E_k} = \frac{3}{2}kT\right) \), so as \( T \) rises from 290 K to 320 K, the mean kinetic energy increases proportionally (by a factor of \( 320/290 \), i.e. from \( 6.00\times10^{-21}\ \text{J} \) to \( 6.62\times10^{-21}\ \text{J} \)).

Final answer: \( p_2 = 1.49\times10^{5}\ \text{Pa} \).

(a) [1] correct statement/use of combined gas law \( p_1V_1/T_1=p_2V_2/T_2 \); [1] correct substitution; [1] \( p_2 = 1.49\times10^{5}\ \text{Pa} \) shown to at least 3 s.f. supporting the given value (reject if working not shown for a 'show that'). (b) [1] correct substitution into \( n=p_1V_1/(RT_1) \); [1] \( n = 0.101\ \text{mol} \) (accept 0.100–0.102). (c) [1] use of \( N=nN_A \); [1] \( N = 6.06\times10^{22} \) (ECF from (b)). (d) [1] correct formula \( \frac32 kT \); [1] correct substitution; [1] \( 6.00\times10^{-21}\ \text{J} \) (accept 5.95–6.05 x10^-21). (e) [1] mean k.e. increases; [1] correct reasoning that mean k.e. is proportional to absolute temperature; [1] supporting numerical comparison or correct new value \( \approx 6.62\times10^{-21}\ \text{J} \); reject answers referring to Celsius temperature.

(a) Two forces: weight \( mg \) acting vertically downward at the ball, and tension \( T \) acting along the string toward the fixed point.

(b) \( r = L\sin\theta = 0.800 \times \sin(25.0^\circ) = 0.338\ \text{m} \)

(c) Vertically: \( T\cos\theta = mg \Rightarrow T = \frac{mg}{\cos\theta} = \frac{0.150\times9.81}{\cos(25.0^\circ)} = 1.62\ \text{N} \)

(d) Horizontally, the resultant (unbalanced) force provides the centripetal force: \( T\sin\theta = \frac{mv^2}{r} \Rightarrow v = \sqrt{\frac{T\sin\theta \times r}{m}} = \sqrt{\frac{(1.62)(\sin25.0^\circ)(0.338)}{0.150}} = 1.24\ \text{m s}^{-1} \)

(e) \( v = \frac{2\pi r}{T_{period}} \Rightarrow T_{period} = \frac{2\pi r}{v} = \frac{2\pi(0.338)}{1.24} = 1.71\ \text{s} \)

Final answer: time for one revolution \( = 1.71\ \text{s} \).

(a) [1] weight vertically down at ball; [1] tension along string toward the fixed support point (reject a separate 'centripetal force' shown as a third arrow). (b) [1] \( r=L\sin\theta = 0.338\ \text{m} \) shown. (c) [1] correct vertical resolution \( T\cos\theta=mg \); [1] correct rearrangement; [1] \( T=1.62\ \text{N} \) shown to 3 s.f. (d) [1] correct horizontal equation \( T\sin\theta = mv^2/r \); [1] correct rearrangement for v; [1] \( v = 1.24\ \text{m s}^{-1} \) (accept 1.23–1.25). (e) [1] use of \( v=2\pi r/T_{period} \) or \( \omega=v/r \) then \( T_{period}=2\pi/\omega \); [1] correct substitution; [1] \( T_{period}=1.71\ \text{s} \) (ECF from (d)).

(a) Simple harmonic motion is oscillatory motion in which the acceleration is directly proportional to the displacement from a fixed (equilibrium) point, and is always directed towards that point (i.e. \( a=-\omega^2 x \)).

(b) \( \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{18.0}{0.250}} = 8.49\ \text{rad s}^{-1} \)

(c) \( T = \frac{2\pi}{\omega} = \frac{2\pi}{8.49} = 0.740\ \text{s} \)

(d) \( v_{max} = \omega A = 8.49 \times 0.060 = 0.509\ \text{m s}^{-1} \)

(e) \( v = \omega\sqrt{A^2-x^2} = 8.49\sqrt{0.060^2-0.030^2} = 0.441\ \text{m s}^{-1} \)

(f) \( a_{max} = \omega^2 A = 8.49^2 \times 0.060 = 4.32\ \text{m s}^{-2} \)

Final answer: period \( T = 0.740\ \text{s} \).

(a) [1] acceleration proportional to displacement from a fixed point; [1] directed towards that point / opposite direction to displacement (allow \( a=-\omega^2x \) with terms defined for both marks). (b) [1] correct substitution into \( \omega=\sqrt{k/m} \); [1] \( \omega=8.49\ \text{rad s}^{-1} \) shown. (c) [1] use of \( T=2\pi/\omega \); [1] \( T=0.740\ \text{s} \) (accept 0.739–0.741). (d) [1] use of \( v_{max}=\omega A \); [1] \( 0.509\ \text{m s}^{-1} \). (e) [1] correct substitution into \( v=\omega\sqrt{A^2-x^2} \); [1] \( 0.441\ \text{m s}^{-1} \) (accept 0.44). (f) [1] use of \( a_{max}=\omega^2A \); [1] \( 4.32\ \text{m s}^{-2} \) (ECF from (b)).

(a) Activity is the number of nuclear disintegrations occurring per unit time (i.e. the rate at which nuclei decay), \( A=-\lambda N \); unit becquerel, Bq (1 Bq = 1 decay per second).

(b) \( \lambda = \frac{\ln2}{t_{1/2}} = \frac{\ln2}{6.0\times3600} = \frac{0.693}{21600} = 3.21\times10^{-5}\ \text{s}^{-1} \)

(c) \( A_0 = \lambda N_0 = (3.21\times10^{-5})(4.80\times10^{12}) = 1.54\times10^{8}\ \text{Bq} \)

(d) \( A = A_0 e^{-\lambda t} \), with \( t = 15.0\times3600 = 5.40\times10^4\ \text{s} \): \( A = 1.54\times10^{8} \times e^{-(3.21\times10^{-5})(5.40\times10^4)} = 1.54\times10^{8} \times e^{-1.733} = 2.72\times10^{7}\ \text{Bq} \)

(e) The statement is correct. Half-life is constant regardless of the amount of the isotope remaining, because radioactive decay is a random and exponential process where the decay constant \( \lambda \) (the probability of decay per nucleus per second) does not change with time. So, however much activity remains after any point, it will always halve again after a further 6.0 hours — not 3.0 hours. The technician's statement (halving in 3.0 hours) is therefore incorrect: halving always takes one full half-life of 6.0 hours, independent of how many half-lives have already elapsed.

Final answer: \( A(15\ \text{h}) = 2.72\times10^{7}\ \text{Bq} \).

(a) [1] rate of decay/disintegration of nuclei (allow \( A=\lambda N \)); [1] becquerel, Bq. (b) [1] correct substitution of \( t_{1/2} \) in seconds into \( \lambda=\ln2/t_{1/2} \); [1] \( \lambda=3.21\times10^{-5}\ \text{s}^{-1} \) shown. (c) [1] correct substitution into \( A_0=\lambda N_0 \); [1] \( A_0=1.54\times10^{8}\ \text{Bq} \). (d) [1] correct use of \( A=A_0e^{-\lambda t} \) with t converted to seconds; [1] correct substitution; [1] \( A=2.72\times10^{7}\ \text{Bq} \) (accept 2.6–2.8 x10^7, ECF from (c)). (e) Mark as a judgement — allow full credit for a well-reasoned 'incorrect' conclusion: [1] correctly identifies that half-life is independent of the quantity of isotope remaining / constant for a given isotope; [1] correct reasoning that decay is exponential/random with constant \( \lambda \); [1] concludes the statement is incorrect because halving always requires 6.0 hours, not 3.0 hours, however many half-lives have passed; reject answers that state activity decreases linearly.

(a) Nucleon number (mass number), A, is the total number of protons and neutrons in a nucleus. For radon-222, \( A = 222 \).

(b) Both sides of the decay equation are neutral atoms (Ra has 88 electrons; Rn has 86 electrons and the emitted \( ^4\text{He} \) atom has 2 electrons, and \( 86+2=88 \)), so the number of electrons is the same on both sides and their masses cancel; using atomic masses therefore gives the same mass defect as using purely nuclear masses.

(c) \( \Delta m = m(^{226}\text{Ra}) - \left[m(^{222}\text{Rn})+m(^{4}\text{He})\right] = 226.025410 - (222.017578+4.002602) = 0.005230\ \text{u} \)
\( \Delta m = 0.005230 \times 1.66\times10^{-27} = 8.68\times10^{-30}\ \text{kg} \)

(d) \( E = \Delta m c^2 = (8.68\times10^{-30})(3.00\times10^{8})^2 = 7.81\times10^{-13}\ \text{J} \)
In MeV: \( E = \frac{7.81\times10^{-13}}{1.60\times10^{-19}\times10^{6}} = 4.88\ \text{MeV} \)

(e) The kinetic energy of the emitted alpha particle and the recoiling radon-222 nucleus (energy may also appear as a gamma-ray photon if the radon nucleus is left in an excited state).

Final answer: \( E \approx 4.88\ \text{MeV} \) released.

(a) [1] total number of protons and neutrons in the nucleus; [1] \( A=222 \). (b) [1] identifies that electron numbers balance on both sides (88 = 86 + 2); [1] therefore electron masses cancel and atomic masses give the correct mass defect. (c) [1] correct subtraction giving \( \Delta m = 0.005230\ \text{u} \) (accept 0.00523); [1] correct conversion factor \( 1.66\times10^{-27}\ \text{kg/u} \) used; [1] \( \Delta m = 8.68\times10^{-30}\ \text{kg} \) (accept 8.6–8.7 x10^-30). (d) [1] correct use of \( E=\Delta mc^2 \); [1] \( E=7.81\times10^{-13}\ \text{J} \) (ECF from (c)); [1] correct conversion using \( 1\ \text{eV}=1.60\times10^{-19}\ \text{J} \); [1] \( E=4.88\ \text{MeV} \) (accept 4.8–4.9 MeV). (e) [1] kinetic energy of the alpha particle and/or recoiling nucleus; [1] second valid form, e.g. gamma photon energy if daughter nucleus de-excites; reject 'heat' alone with no reference to KE of the fragments.

An indicative full-mark response would include the following points, in a well-organised, clearly-written answer:

Binding energy per nucleon: the binding energy per nucleon curve rises steeply for light nuclei, peaks around iron (nucleon number about 56), and falls slowly for heavy nuclei. Both fission of very heavy nuclei (e.g. uranium-235) and fusion of very light nuclei (e.g. deuterium and tritium) move nuclei towards the peak of the curve, increasing the binding energy per nucleon and releasing energy consistent with \( E=\Delta mc^2 \).

Fission reactor: a neutron is absorbed by a uranium-235 nucleus, causing it to split into two smaller daughter nuclei plus typically two or three fast neutrons and a release of energy. If, on average, one of these neutrons goes on to cause a further fission, a self-sustaining chain reaction results; the critical size is the minimum mass/size of fuel for which enough neutrons are retained (rather than escaping) to sustain the chain reaction. A moderator (e.g. graphite or water) slows fast neutrons to thermal speeds, at which they are far more likely to cause further fission in U-235. Control rods (e.g. boron or cadmium) absorb neutrons and can be inserted or withdrawn to regulate the reaction rate and hold it at a steady, critical level. A cooling system removes the heat produced (transferring it, ultimately, to drive turbines and generators) and reactor shielding (e.g. thick concrete and steel) absorbs radiation to protect workers and the environment.

Fusion (ITER/tokamak): the fuel is a mixture of deuterium and tritium (both isotopes of hydrogen), which fuse via \( ^2_1\text{H} + \,^3_1\text{H} \rightarrow \,^4_2\text{He} + \,^1_0\text{n} \), releasing energy. This reaction requires extremely high temperatures (of the order of \( 10^8\ \text{K} \)) for the positively-charged nuclei to overcome their electrostatic (Coulomb) repulsion and get close enough to fuse. At these temperatures the fuel exists as a plasma (an ionised gas of nuclei and free electrons). Plasma heating methods include ohmic heating (from the induced plasma current), neutral beam injection, and radio-frequency/microwave heating. Because no solid vessel wall can withstand contact with plasma at these temperatures, the hot plasma is confined away from the walls of the vacuum vessel using strong magnetic fields (magnetic confinement) shaped by the tokamak's toroidal and poloidal field coils. A surrounding 'blanket' is designed to absorb neutron energy (and, in future power reactors, to breed further tritium fuel).

Advantages/disadvantages and social, environmental, security and economic issues: Fission is a proven, reliable technology that can supply continuous, high-density energy without direct greenhouse-gas emissions during operation, but it produces long-lived radioactive waste that must be safely stored for very long periods, carries a risk (however small) of serious accident, and raises proliferation/security concerns because reactor-grade material and technology could potentially be diverted towards weapons use; reactors are also very expensive to build and eventually decommission. Fusion fuel (deuterium from water and tritium bred from lithium) is abundant and produces little long-lived radioactive waste and no possibility of a runaway chain reaction (if confinement fails, the plasma simply cools and the reaction stops), making it inherently safer and more sustainable; however, achieving and maintaining the extreme temperatures and stable magnetic confinement needed for a sustained, net-energy-positive reaction on a practical, terrestrial scale is an enormous and not yet fully solved engineering challenge, and no commercial fusion power station currently exists, so very large research and development costs must be met before fusion could contribute to solving the energy crisis.

This is a levels-of-response (QWC) question marked holistically out of 25 marks against scientific content and quality of written communication together.

Level 3 (17–25 marks): Detailed, accurate and well-organised account covering the physical principles of BOTH fission and fusion in appropriate depth (binding energy curve; chain reaction/critical size/moderator/control rods/cooling for fission; D-T reaction/temperature/plasma/heating methods/magnetic confinement for fusion) AND a balanced discussion of advantages/disadvantages/social/environmental/security/economic issues for both. Correct use of specialist terminology throughout; answer is coherently structured with accurate spelling, punctuation and grammar.

Level 2 (9–16 marks): Reasonably accurate description of the principles of fission and fusion, though one process may be covered in less depth, or with a small number of errors/omissions among the required terms (e.g. missing critical size or plasma heating methods); discussion of issues present but less developed on one side. Mostly correct terminology; generally clear organisation with occasional lapses in grammar or structure.

Level 1 (1–8 marks): Limited, fragmented or largely one-sided account (e.g. describes only fission, or only in vague/non-technical terms); little or no discussion of advantages/disadvantages/wider issues; frequent errors in terminology or communication that hinder understanding.

0 marks: No creditable response, or answer contains no relevant physics.

Award marks within each band according to the completeness, accuracy and balance of coverage of fission AND fusion, and the clarity/coherence of the writing. Full credit at Level 3 requires substantive, correct content on both fission and fusion, not just one.

(a) Gravitational field strength at a point is the gravitational force exerted per unit mass on a small test mass placed at that point, \( g=F/m \).

(b) Every point mass attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

(c) \( g = \frac{GM}{R^2} = \frac{(6.67\times10^{-11})(6.40\times10^{23})}{(3.40\times10^{6})^2} = 3.69\ \text{m s}^{-2} \)

(d) For circular orbital motion, the gravitational force provides the centripetal force: \( \frac{GMm}{r^2} = \frac{mv^2}{r} \). Dividing both sides by \( m \) and multiplying by \( r \): \( \frac{GM}{r} = v^2 \), so \( v=\sqrt{\frac{GM}{r}} \).
\( v = \sqrt{\frac{(6.67\times10^{-11})(6.40\times10^{23})}{9.40\times10^{6}}} = 2.13\times10^{3}\ \text{m s}^{-1} \)

(e) \( T = \frac{2\pi r}{v} = \frac{2\pi(9.40\times10^{6})}{2.13\times10^{3}} = 2.77\times10^{4}\ \text{s} = 7.70\ \text{hours} \)

(f) The orbital period would increase. From Kepler's third law (\( T^2 \propto r^3 \), consistent with Newton's law of gravitation), a larger orbital radius requires a longer period; physically, at larger r the gravitational field strength (and hence the required centripetal acceleration) is smaller, so the probe needs a lower speed and takes longer to complete each orbit.

Final answer: \( v = 2.13\times10^{3}\ \text{m s}^{-1} \).

(a) [1] force per unit mass; [1] acting on a small/test mass at that point. (b) [1] force proportional to product of the masses; [1] inversely proportional to square of separation (between centres of mass). (c) [1] correct substitution into \( g=GM/R^2 \); [1] correct powers of ten; [1] \( g=3.69\ \text{m s}^{-2} \) (accept 3.65–3.75). (d) [1] correct equation \( GMm/r^2 = mv^2/r \); [1] correct algebraic cancellation of m and one r shown; [1] correct substitution; [1] \( v=2.13\times10^{3}\ \text{m s}^{-1} \) (accept 2.10–2.16 x10^3). (e) [1] correct use of \( T=2\pi r/v \); [1] \( T=7.70\ \text{hours} \) (accept 7.6–7.8 h; ECF from (d)). (f) [1] period increases; [1] correct physical reasoning referencing \( T^2\propto r^3 \) or reduced field strength/required speed at larger r; reject 'period increases' with no valid reasoning.

(a) The force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them: \( F = \dfrac{q_1q_2}{4\pi\varepsilon_0 r^2} \).

(b) \( F = \frac{kq_1q_2}{r^2} = \frac{(8.99\times10^{9})(3.20\times10^{-9})(5.00\times10^{-9})}{(0.150)^2} = 6.39\times10^{-6}\ \text{N} \). Since the charges are of opposite sign, the force is attractive.

(c) \( E = \frac{kq_A}{r^2} = \frac{(8.99\times10^{9})(3.20\times10^{-9})}{(0.150)^2} = 1.28\times10^{3}\ \text{N C}^{-1} \), directed away from A (towards B).

(d) No. Although the test charge is equidistant from A and B, A and B carry charges of different sign and different magnitude, so the field contributions from A and B at the midpoint are neither equal nor opposite in effect; both fields point in the same direction at the midpoint (away from positive A and towards negative B), so they add rather than cancel, giving a non-zero resultant force.

(e) Similarity: both fields obey an inverse-square law with distance (field strength \( \propto 1/r^2 \)), and both are radial, central fields around a point source. Difference: the electric field can be either attractive or repulsive (like charges repel, unlike attract) depending on the sign of the charges, whereas the gravitational field is always attractive (mass is never negative).

(f) The field strength is largest close to A and decreases with distance according to an inverse-square relationship, tending towards zero as \( r\rightarrow\infty \), but never reaching exactly zero at finite r.

Final answer: \( F = 6.39\times10^{-6}\ \text{N} \), attractive.

(a) [1] force proportional to product of charges; [1] inversely proportional to square of separation (accept the formula with terms defined as an alternative to prose). (b) [1] correct substitution; [1] \( F=6.39\times10^{-6}\ \text{N} \) (accept 6.3–6.5 x10^-6); [1] correctly states attractive (with reason: opposite signs). (c) [1] correct substitution into \( E=kq/r^2 \); [1] \( E=1.28\times10^{3}\ \text{N C}^{-1} \) (accept 1.25–1.31 x10^3); [1] correct direction (away from A, i.e. towards B). (d) [1] correctly identifies fields do not cancel (states 'no'); [1] valid reasoning referring to unequal charge magnitudes and/or fields acting in the same direction at the midpoint. (e) [1] valid similarity (inverse-square law and/or radial field); [1] valid difference (electric field can be attractive or repulsive vs gravity always attractive); [1] for a second correctly explained point (e.g. electric field depends on charge, gravitational on mass) OR a clear, fully-justified single pair of points; award up to 3 marks for any two correct, clearly distinguished similarity/difference statements. (f) [1] field strength greatest near A, decreasing with r; [1] correct inverse-square (non-linear) shape described, approaching but not reaching zero.

(a) Magnetic flux linkage of a coil is the product of the number of turns of the coil and the magnetic flux passing through it, \( N\Phi = NBA \).

(b) The magnitude of the induced e.m.f. is directly proportional to the rate of change of magnetic flux linkage.

(c) \( \varepsilon = \left|\frac{N\Delta\Phi}{\Delta t}\right| = \frac{N A \Delta B}{\Delta t} = \frac{250 \times (4.50\times10^{-3}) \times (0.080-0.020)}{0.150} = 0.450\ \text{V} \)

(d) Lenz's law states that the direction of an induced e.m.f. (and hence induced current) is always such as to oppose the change producing it. Here, the flux through the coil is decreasing, so the induced current flows in the direction that would, by its own magnetic field, try to maintain (reinforce) the original flux — i.e. the induced current creates a magnetic field in the same direction as the decreasing field, opposing the decrease.

(e) \( \frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow V_s = V_p \times \frac{N_s}{N_p} = 230 \times \frac{60}{1200} = 11.5\ \text{V} \)

(f) For an ideal transformer, input power = output power: \( V_pI_p = V_sI_s \Rightarrow I_s = \frac{V_pI_p}{V_s} = \frac{230\times0.50}{11.5} = 10.0\ \text{A} \) (equivalently, \( I_s=I_p\times N_p/N_s = 0.50\times1200/60=10.0\ \text{A} \)).

Final answer: \( \varepsilon = 0.450\ \text{V} \).

(a) [1] product of number of turns and flux; [1] correct symbol/relation \( N\Phi \) or \( NBA \). (b) [1] induced e.m.f. proportional to rate of change of flux linkage — must reference 'rate of change' for the second mark. (c) [1] correct use of \( \varepsilon=N\Delta\Phi/\Delta t \) (or via \( \Delta(NBA)/\Delta t \)); [1] correct substitution; [1] \( \varepsilon=0.450\ \text{V} \). (d) [1] correct statement of Lenz's law (opposes the change that causes it); [1] identifies flux is decreasing; [1] correctly explains induced current direction acts to maintain/reinforce the original field (oppose the decrease). (e) [1] correct substitution into transformer equation; [1] \( V_s=11.5\ \text{V} \). (f) [1] correct statement/use of power conservation \( V_pI_p=V_sI_s \) (or turns-ratio current relation); [1] correct substitution; [1] \( I_s=10.0\ \text{A} \) (ECF from (e)).

(a) Capacitance is the charge stored per unit potential difference across a capacitor, \( C=Q/V \); unit farad, F.

(b) \( E = \frac12 CV^2 = \frac12 (470\times10^{-6})(9.00)^2 = 1.90\times10^{-2}\ \text{J} \) (19.0 mJ)

(c) The time constant, \( \tau \), is the time taken for the charge (or p.d., or current) in a discharging RC circuit to fall to \( 1/e \) (about 37%) of its initial value; \( \tau = RC \). \( \tau = (2.2\times10^{3})(470\times10^{-6}) = 1.03\ \text{s} \)

(d) \( V = V_0 e^{-t/\tau} = 9.00 \times e^{-2.0/1.03} = 1.30\ \text{V} \)

(e) The graph of charge against time is an exponential decay curve: it starts at the initial charge \( Q_0=CV_0 \), falls steeply at first, and the rate of fall continually decreases so the curve approaches (but never quite reaches) the time axis; equal time intervals correspond to equal fractional decreases in charge.

(f) A capacitor can be charged slowly (over a relatively long time, limited by the battery's internal resistance/maximum current) but store a large amount of energy \( \left(E=\frac12 CV^2\right) \); when discharged through a low-resistance load (flash tube / defibrillator paddles), the small time constant \( \tau=RC \) of that discharge path allows nearly all of the stored energy to be delivered in a very short burst, giving a much higher instantaneous power than the battery could supply directly on its own.

Final answer: \( V(2.0\,\text{s}) = 1.30\ \text{V} \).

(a) [1] charge stored per unit p.d. / \( C=Q/V \); [1] unit farad (F). (b) [1] correct substitution into \( \frac12CV^2 \); [1] \( E=1.90\times10^{-2}\ \text{J} \) (accept 19.0 mJ). (c) [1] correct definition referencing fall to 1/e (~37%) of initial value; [1] \( \tau=RC \) stated/used; [1] \( \tau=1.03\ \text{s} \). (d) [1] correct use of \( V=V_0e^{-t/\tau} \); [1] correct substitution; [1] \( V=1.30\ \text{V} \) (accept 1.25–1.35, ECF from (c)). (e) [1] correct exponential decay shape described (steep initial fall, levelling off, never reaching zero); [1] identifies it as an exponential (decay) curve. (f) [1] identifies capacitor stores large energy \( \frac12CV^2 \) while being charged slowly/at low current; [1] identifies the discharge time constant through the low-resistance load is very small; [1] correctly links this to a large power / rapid energy delivery not achievable directly from the battery.

(a) \( E_0 = m_ec^2 = (9.11\times10^{-31})(3.00\times10^{8})^2 = 8.20\times10^{-14}\ \text{J} = \dfrac{8.20\times10^{-14}}{1.60\times10^{-19}\times10^6} = 0.512\ \text{MeV} \)

(b) As a particle's speed approaches the speed of light, its total energy (and hence its effective inertial mass, \( m=\gamma m_0 \)) increases without limit, so ever-larger forces are needed to produce the same acceleration. This relativistic mass increase only becomes significant when the particle's speed is an appreciable fraction of the speed of light \( c \) (i.e. in the relativistic regime, \( v \) not \( \ll c \)).

(c) Total energy = rest energy + kinetic energy: \( \gamma m_ec^2 = m_ec^2 + KE \Rightarrow \gamma = 1+\dfrac{KE}{m_ec^2} = 1+\dfrac{2.0}{0.512} = 4.90 \)

(d) \( \gamma=\dfrac{1}{\sqrt{1-v^2/c^2}} \Rightarrow 1-\dfrac{v^2}{c^2}=\dfrac{1}{\gamma^2} \Rightarrow \dfrac{v}{c}=\sqrt{1-\dfrac{1}{\gamma^2}} = \sqrt{1-\dfrac{1}{4.90^2}} = 0.979 \), so \( v \approx 0.979c \ (2.94\times10^{8}\ \text{m s}^{-1}) \).

(e) Antimatter consists of antiparticles: particles with the same mass as their corresponding ordinary matter particle but with opposite charge (and opposite values of other properties such as lepton/baryon number). In a high-energy accelerator, when particles are collided at sufficient energy, some of the kinetic energy can be converted into mass, creating particle–antiparticle pairs (e.g. an electron and a positron), consistent with \( E=mc^2 \).

(f) When an electron and a positron meet, they annihilate: their combined rest mass and kinetic energy are converted entirely into electromagnetic radiation, typically two gamma-ray photons emitted in (approximately) opposite directions (two photons, rather than one, are needed so that momentum, which was approximately zero before the annihilation, is also conserved). The quantities conserved in the interaction are: energy, (linear) momentum, and charge (total charge is zero both before and after).

Final answer: \( \gamma \approx 4.90 \).

(a) [1] correct substitution into \( E_0=m_ec^2 \); [1] correct conversion to MeV giving \( \approx0.51\ \text{MeV} \), shown to at least 2 s.f. supporting the given value. (b) [1] correct description of increasing effective mass/inertia with speed (\( m=\gamma m_0 \)); [1] correctly states this is significant only as v approaches c (relativistic speeds). (c) [1] correct relation total energy = rest energy + KE; [1] correct rearrangement for \( \gamma \); [1] \( \gamma\approx4.9 \) shown (accept 4.85–4.95, ECF from (a)). (d) [1] correct rearrangement of the Lorentz factor equation for v/c; [1] correct substitution; [1] \( v/c=0.979 \) or \( v=2.94\times10^{8}\ \text{m s}^{-1} \) (accept 0.975–0.983, ECF from (c)). (e) [1] correct description of antiparticles (same mass, opposite charge/quantum numbers); [1] correctly states production via conversion of kinetic energy to mass in high-energy collisions (pair production). (f) [1] correctly identifies emission of (two) gamma-ray photons; [1] correct reasoning that two photons (in opposite directions) are required to conserve momentum; [1] correctly states energy, momentum and charge are conserved (any two of these for full credit if clearly stated, allow 1 mark for one only).

(a) A fundamental particle is one with no internal substructure — it is not made up of any smaller, simpler particles (e.g. quarks and leptons, as far as is currently known).

(b) The electron and the antineutrino are both leptons (fundamental particles which do not experience the strong nuclear interaction). The proton and neutron are both hadrons, and specifically baryons (particles made of three quarks).

(c) Proton: two up quarks and one down quark (uud). Neutron: one up quark and two down quarks (udd).

(d) In \( \beta^- \) decay, one of the down quarks within the neutron (udd) changes into an up quark, converting the neutron (udd) into a proton (uud); a \( W^- \) boson is emitted in this quark transition, which itself immediately decays into an electron and an antineutrino ( \( W^- \rightarrow e^- + \bar\nu_e \) ). The exchange particle responsible is the \( W^- \) boson, and the fundamental interaction involved is the weak (nuclear) interaction.

(e) Before: charge of neutron \( = 0 \). After: charge of proton \( (+1e) \) + charge of electron \( (-1e) \) + charge of antineutrino \( (0) \) \( = +1e-1e+0 = 0 \). Since \( 0=0 \), charge is conserved.

(f) Baryon number: neutron has baryon number \( +1 \) (it is a baryon). After decay: proton has baryon number \( +1 \); the electron and antineutrino are leptons, not baryons, so each has baryon number \( 0 \). Total after \( = 1+0+0 = 1 \). Since before \( =1 \) and after \( =1 \), baryon number is conserved.

Final answer: a down quark in the neutron converts into an up quark, emitting a \( W^- \) boson which decays to \( e^-+\bar\nu_e \); the interaction is the weak interaction.

(a) [1] particle with no substructure / not composed of smaller particles. (b) [1] electron and antineutrino both leptons; [1] proton and neutron both hadrons; [1] specifically baryons. (c) [1] proton = uud; [1] neutron = udd. (d) [1] identifies a down quark converts to an up quark; [1] correctly states this occurs within the neutron, converting udd to uud; [1] identifies the \( W^- \) boson as the exchange particle (and that it decays to \( e^-+\bar\nu_e \)); [1] correctly identifies the weak interaction as the interaction responsible. (e) [1] correct statement of charges before and after; [1] correctly shows \( 0=(+1)+(-1)+0 \), charge conserved. (f) [1] correctly assigns baryon number +1 to neutron and proton, 0 to electron and antineutrino; [1] correct total before (1) and after (1); [1] correctly concludes baryon number is conserved.

An indicative full-mark response would include:

Electric field: a charge in a uniform electric field experiences a force \( F=qE \), which acts in the direction of the field (for a positive charge) or opposite to the field (for a negative charge, such as the electron here); the force is constant in magnitude and direction throughout the field, since E is uniform and does not depend on the electron's speed. This constant force, perpendicular to the electron's initial velocity, produces a constant acceleration perpendicular to the initial motion, analogous to projectile motion; the resulting path is a parabola.

Magnetic field: a moving charge in a uniform magnetic field perpendicular to its velocity experiences a force \( F=Bqv \), whose direction is given by \( \vec{F}=q\vec{v}\times\vec{B} \) (right-hand/left-hand rule depending on charge sign) and is always perpendicular to both the velocity and the field, not simply along the field direction. Because the force is always perpendicular to the velocity, it does no work on the electron, so the electron's speed (and hence the magnitude of the force, \( F=Bqv \)) stays constant, but its direction continuously changes; this is precisely the condition for circular motion, so the electron moves in a circular arc, with the magnetic force providing the centripetal force.

Comparison: both forces act on the (charged) electron and cause deflection from its original straight-line path; both magnitudes depend on the size of the charge. However, the electric force is independent of speed and produces a parabolic path (as in projectile motion, with the force constant in magnitude and direction), whereas the magnetic force depends on speed (F=Bqv, zero if v=0) and is always perpendicular to the velocity, changing direction as the electron moves and producing a circular path rather than a parabolic one.

Levels-of-response (QWC) mark scheme, out of 10 marks.

Level 3 (7–10 marks): Clear, accurate and well-organised comparison covering: correct force equations \( F=qE \) and \( F=Bqv \) with correct dependence identified (E-force independent of speed; B-force dependent on speed and zero at v=0); correct force directions (E-force along/against field; B-force perpendicular to both v and B); and correctly identifies parabolic path in the electric field versus circular path in the magnetic field, with valid physical reasoning (constant perpendicular force / analogy with projectiles for E; force always perpendicular to v, doing no work, constant speed, centripetal force for B). Correct specialist terminology, coherent structure, accurate spelling, punctuation and grammar.

Level 2 (4–6 marks): Most of the above points made but with some omissions or minor inaccuracies (e.g. correct paths identified but reasoning for one case incomplete, or force direction not fully explained); generally clear communication with occasional lapses.

Level 1 (1–3 marks): Only basic or partial comparison offered (e.g. states the two force equations only, or names the two path shapes without justification); weak use of terminology or poor organisation.

0 marks: No creditable response.

(a) Points should be plotted accurately (within ±1 small square) with F on the x-axis and x on the y-axis, using scales that occupy at least half of the grid in each direction; a single straight line of best fit is drawn through the origin and the plotted points, since \( x = \left(\frac{L}{AE}\right)F \) is a direct proportionality (Hooke's law).

(b) The data are exactly linear: gradient \( = \dfrac{7.31-0.00}{50.0-0.0} = 0.1462\ \text{mm N}^{-1} = 1.462\times10^{-4}\ \text{m N}^{-1} \).

(c) \( A = \pi\left(\dfrac{d}{2}\right)^2 = \pi\left(\dfrac{0.28\times10^{-3}}{2}\right)^2 = 6.16\times10^{-8}\ \text{m}^2 \)

(d) Since \( x=\dfrac{FL}{AE} \), the gradient of the x–F graph, \( m=\dfrac{x}{F}=\dfrac{L}{AE} \), so \( E = \dfrac{L}{Am} \).
\( E = \dfrac{1.800}{(6.16\times10^{-8})(1.462\times10^{-4})} = 2.00\times10^{11}\ \text{Pa} \)

(e) Precautions: keep the wire straight and under a small initial (verification) tension before taking the zero reading, to remove kinks; take the diameter reading at several points along the wire and average, rotating the micrometer 90° at each point, to allow for non-circular cross-section; view the metre rule scale at eye level (avoid parallax error) when reading the extension; add/remove loads gently to avoid sudden jerks that could exceed the elastic limit (any two valid, distinct precautions).

(f) A travelling microscope (or a vernier scale/reference pointer with vernier calipers) could be used to measure the extension/length more precisely than a metre rule, because it has a much finer resolution (typically 0.01 mm or better, compared with about 1 mm for a metre rule), reducing the percentage uncertainty in the length and extension measurements.

(g) A thicker wire has a larger cross-sectional area A. Since the gradient \( m=L/(AE) \) is inversely proportional to A (for the same L and E), a thicker wire of the same material and length would give a smaller gradient (a shallower extension–load graph), i.e. it would extend less for the same load.

Final answer: \( E = 2.00\times10^{11}\ \text{Pa} \) (200 GPa).

(a) [1] suitable linear scales using at least half the grid; [1] all 6 points plotted accurately (±1 small square); [1] single straight best-fit line, not a dot-to-dot join; [1] line passes through/near the origin consistent with the data. (b) [1] triangle/points used span at least half the line drawn; [1] correct gradient calculation shown; [1] \( m=1.46\times10^{-4}\ \text{m N}^{-1} \) (accept 1.44–1.48 x10^-4) with correct unit. (c) [1] correct substitution into \( A=\pi(d/2)^2 \); [1] \( A=6.16\times10^{-8}\ \text{m}^2 \). (d) [1] correct rearrangement \( E=L/(Am) \) (or equivalent via stress/strain from gradient) shown clearly; [1]–[3] correct substitution and calculation stages; [1] \( E=2.00\times10^{11}\ \text{Pa} \) (accept 1.90–2.10 x10^11, ECF from (b) and (c)); reject missing/incorrect unit. (e) [1] each for any two valid, distinct precautions (as in solution or equivalent, e.g. avoiding temperature changes/draughts, ensuring wire not overloaded past elastic limit); max [2]. (f) [1] valid higher-precision instrument named (travelling microscope / vernier scale); [1] correct reasoning linking finer resolution to reduced percentage uncertainty. (g) [1] correctly identifies gradient would decrease (thicker wire extends less); [1] correct reasoning via inverse proportionality between gradient and cross-sectional area.

(a) Taking natural logs of \( V=V_0e^{-t/RC} \): \( \ln V = \ln V_0 - \dfrac{1}{RC}t \). This is of the form \( y=mx+c \) with \( y=\ln V \), \( x=t \), gradient \( m=-\dfrac{1}{RC}=-\dfrac{1}{\tau} \), and y-intercept \( c=\ln V_0 \) (the natural log of the initial p.d.). So a graph of \( \ln V \) against t should be a straight line, with gradient equal to \( -1/\tau \) and y-intercept equal to \( \ln V_0 \).

(b) Points plotted accurately using scales occupying at least half the grid; single straight best-fit line drawn.

(c) The data are exactly linear: gradient \( = \dfrac{0.592-1.792}{12.0-0.0} = \dfrac{-1.200}{12.0} = -0.1000\ \text{s}^{-1} \).

(d) Since gradient \( =-1/\tau \): \( \tau = \dfrac{-1}{\text{gradient}} = \dfrac{-1}{-0.1000} = 10.0\ \text{s} \).

(e) \( \tau=RC \Rightarrow C=\dfrac{\tau}{R}=\dfrac{10.0}{1.00\times10^{5}}=1.00\times10^{-4}\ \text{F} =100\ \mu\text{F} \). This agrees exactly with the capacitor's stated value of \( 100\ \mu\text{F} \), confirming the component is within its stated tolerance.

(f) A systematic error could arise from the resistance of the voltmeter itself (a real voltmeter is not infinite resistance, so it draws some current and provides an additional discharge path in parallel with R); this would make the true effective resistance of the discharge path slightly less than the quoted R, causing the capacitor to discharge slightly faster than expected and giving a measured \( \tau \) that is slightly smaller than the true value. (Alternative valid answers: reaction-time delay in starting the stopwatch at the instant of switching would shift all readings and could bias the intercept but not the gradient/\( \tau \) itself; a systematic zero-error in the voltmeter would bias V0 similarly.)

Final answer: \( \tau = 10.0\ \text{s} \).

(a) [1] correct log manipulation of \( V=V_0e^{-t/RC} \) to linear form; [1] correctly identifies y=ln V, x=t; [1] gradient \( =-1/RC(=-1/\tau) \); [1] y-intercept \( =\ln V_0 \). (b) [1] suitable linear scales using at least half the grid; [1] all 7 points plotted accurately (±1 small square); [1] single straight best-fit line; [1] line drawn through/consistent with the full data range. (c) [1] triangle/points used span at least half the line; [1] correct gradient calculation shown; [1] gradient \( =-0.100\ \text{s}^{-1} \) (accept -0.098 to -0.102) with correct unit. (d) [1] correct relation \( \tau=-1/\text{gradient} \) used; [1] correct substitution; [1] \( \tau=10.0\ \text{s} \) (ECF from (c)). (e) [1] correct use of \( C=\tau/R \); [1] \( C=1.00\times10^{-4}\ \text{F}\ (100\ \mu\text{F}) \); [1] valid, correctly-reasoned comparison with the stated value. (f) [1] identifies a genuine systematic (not random) error source; [1] correct physical reasoning for its origin; [1] correctly states the direction of its effect on the calculated \( \tau \).

(a) \( \%\Delta d = \dfrac{0.01}{0.28}\times100 = 3.57\% \). Since \( A\propto d^2 \), \( \%\Delta A = 2\times3.57\% = 7.14\% \).

(b) \( \%\Delta x = \dfrac{0.05}{1.46}\times100 = 3.42\% \); \( \%\Delta L = \dfrac{2}{1800}\times100 = 0.11\% \).

(c) Since \( E=FL/(Ax) \) is a product/quotient of measured quantities, the percentage uncertainties add: \( \%\Delta E = \%\Delta L + \%\Delta A + \%\Delta x = 0.11+7.14+3.42 = 10.7\% \) (F contributes 0%). Absolute uncertainty \( = 2.00\times10^{11}\times0.107 = 2.14\times10^{10}\ \text{Pa} \), so \( E=(2.00\pm0.21)\times10^{11}\ \text{Pa} \).

(d) The diameter measurement contributes the most, because it is doubled (A depends on \( d^2 \)) and already has the largest individual percentage uncertainty (3.57%) of the directly-measured quantities, so its contribution to \( \%\Delta E \) (7.14%) dominates the total.

Final answer: total percentage uncertainty in \( E \approx 10.7\% \), i.e. \( E=(2.00\pm0.21)\times10^{11}\ \text{Pa} \).

(a) [1] \( \%\Delta d=3.57\% \) (accept 3.5–3.6%); [1] correctly doubles to give \( \%\Delta A=7.14\% \). (b) [1] \( \%\Delta x=3.42\% \) (accept 3.4%); [1] \( \%\Delta L=0.11\% \) (accept 0.1%). (c) [1] correctly identifies percentage uncertainties should be summed for a product/quotient formula; [1] \( \%\Delta E\approx10.7\% \) (accept 10.5–11.0%, ECF from (a)/(b)); [1] correct absolute uncertainty \( \approx0.21\times10^{11}\ \text{Pa} \) (ECF). (d) [1] correctly identifies diameter, with valid reasoning referencing the squared dependence and/or its largest individual percentage uncertainty.

(a) \( T=\dfrac{t_{20}}{20}=\dfrac{36.9}{20}=1.845\ \text{s} \); \( \Delta T=\dfrac{\Delta t_{20}}{20}=\dfrac{0.4}{20}=0.020\ \text{s} \). So \( T=(1.845\pm0.020)\ \text{s} \).

(b) \( g=\dfrac{4\pi^2L}{T^2}=\dfrac{4\pi^2(0.850)}{(1.845)^2}=9.86\ \text{m s}^{-2} \)

(c) \( \%\Delta L=\dfrac{0.002}{0.850}\times100=0.24\% \); \( \%\Delta T=\dfrac{0.020}{1.845}\times100=1.08\% \). Since \( g\propto L/T^2 \), \( \%\Delta g=\%\Delta L+2\times\%\Delta T=0.24+2.17=2.40\% \). Absolute uncertainty \( =9.86\times0.0240=0.237\ \text{m s}^{-2} \), so \( g=(9.86\pm0.24)\ \text{m s}^{-2} \).

(d) Timing 20 oscillations means the same fixed reaction-time uncertainty (±0.4 s) is spread over a much longer total time, so once divided by 20 to find T, the absolute uncertainty in T (and hence its percentage uncertainty) is reduced by a factor of 20 compared with timing a single oscillation directly.

Final answer: \( g=(9.86\pm0.24)\ \text{m s}^{-2} \), consistent with the accepted value of \( 9.81\ \text{m s}^{-2} \) within uncertainty.

(a) [1] correct division by 20 for T; [1] \( T=1.845\ \text{s} \); [1] correct \( \Delta T=0.020\ \text{s} \). (b) [1] correct substitution into \( g=4\pi^2L/T^2 \); [1] \( g=9.86\ \text{m s}^{-2} \) (accept 9.80–9.92, ECF from (a)). (c) [1] correct \( \%\Delta L \) and \( \%\Delta T \) calculated; [1] correctly doubles \( \%\Delta T \) and sums with \( \%\Delta L \) to give \( \%\Delta g\approx2.40\% \) (accept 2.3–2.5%); [1] correct absolute uncertainty \( \approx0.24\ \text{m s}^{-2} \) (ECF). (d) [1] correct reasoning that the fixed absolute reaction-time uncertainty, once divided by 20, gives a much smaller uncertainty in T (and hence percentage uncertainty) than timing a single oscillation.

(a) Taking natural logs of \( C=C_0e^{-\lambda t} \): \( \ln C=\ln C_0-\lambda t \), which has the form \( y=mx+c \) with \( y=\ln C \), \( x=t \); gradient \( m=-\lambda \); y-intercept \( c=\ln C_0 \).

(b) \( \ln C \) values: t=0: 6.745; t=20: 6.399; t=40: 6.052; t=60: 5.707 (accept 5.705–5.708); t=80: 5.361 (accept 5.359–5.362); t=100: 5.011 (accept 5.008–5.014); t=120: 4.664 (accept 4.660–4.669).

(c) Points plotted accurately (±1 small square) with suitable linear scales; single straight best-fit line drawn.

(d) The data are exactly linear: gradient \( = \dfrac{4.666-6.745}{120-0} = \dfrac{-2.079}{120} = -0.01733\ \text{s}^{-1} \). Since gradient \( =-\lambda \), \( \lambda=0.0173\ \text{s}^{-1} \).

(e) \( t_{1/2}=\dfrac{\ln2}{\lambda}=\dfrac{0.693}{0.0173}=40.0\ \text{s} \).

(f) The y-intercept is \( \ln C_0=6.745 \), so \( C_0=e^{6.745}=850 \) counts per 10 s (matching the first data point, since t=0 was included in the data range).

(g) Plotting \( \ln C \) against t converts the exponential relationship into a straight line, so a single best-fit line uses ALL of the data points to determine \( \lambda \) (via the gradient), averaging out random errors/scatter across every measurement; reading a half-life directly off the curved C–t graph instead relies on locating just one or two specific points (e.g. where C has halved), which is far more sensitive to error in any single reading and does not make use of all the data.

Final answer: \( \lambda=0.0173\ \text{s}^{-1} \), half-life \( =40.0\ \text{s} \).

(a) [1] correct log manipulation to linear form; [1] gradient \( =-\lambda \); [1] y-intercept \( =\ln C_0 \). (b) [1] for values correct to t=40 inclusive; [1] for values correct t=60 to t=80; [1] for values correct t=100 to t=120 (accept values within the ranges given above, allow rounding to 3 or 4 s.f.). (c) [1] suitable linear scales using at least half the grid; [1] all 7 points plotted accurately (±1 small square, ECF from (b)); [1] single straight best-fit line; [1] line consistent across the full data range. (d) [1] triangle/points used span at least half the line; [1] correct gradient calculation shown; [1] correctly identifies \( \lambda=-\text{gradient} \); [1] \( \lambda=0.0173\ \text{s}^{-1} \) (accept 0.0170–0.0177) with correct unit. (e) [1] correct use of \( t_{1/2}=\ln2/\lambda \); [1] correct substitution; [1] \( t_{1/2}=40.0\ \text{s} \) (accept 38–42 s, ECF from (d)). (f) [1] correctly reads/calculates y-intercept as \( \ln C_0 \); [1] \( C_0\approx850 \) counts per 10 s (accept 800–900, ECF). (g) [1] identifies that the straight-line method uses all data points (via the best-fit line/gradient), reducing the effect of random error in any one reading; [1] identifies that reading directly from a curve relies on one or two points and is more prone to error / harder to draw accurately by eye.

(a) Random error: variation in the student's reaction time each time the stopwatch is started/stopped (causing t20 to vary slightly and unpredictably from repeat to repeat). Systematic error: the length L is measured to the centre of the bob, but if the point of measurement is consistently taken to the top or bottom of the bob instead, every length reading is offset by the same (roughly constant) amount in the same direction.

(b) For the random (timing) error: repeat the timing of \( t_{20} \) several times and take a mean value, which reduces the effect of random reaction-time variations. For the systematic (length) error: use a set-square or fiducial marker aligned with the centre of the bob, and/or use vernier calipers to measure the bob's diameter so that half the diameter can be added to the length to the top of the bob, ensuring L is consistently measured to the bob's centre.

(c) The formula \( T=2\pi\sqrt{L/g} \) (and hence \( g=4\pi^2L/T^2 \)) is only valid for simple harmonic motion, which requires the angular amplitude to be small (conventionally less than about 10°); at 25° the restoring force is no longer accurately proportional to displacement, so the period measured is slightly longer than the small-angle formula predicts, introducing a systematic error into the calculated g (making it come out too low). The experiment should be repeated using a much smaller amplitude (e.g. less than 10°) to remain within the small-angle approximation.

Final answer: main errors are reaction time (random) and offset in measuring L to the bob's centre (systematic); the 25° amplitude introduces a further systematic error and should be reduced to below about 10°.

(a) [1] valid random error (e.g. reaction time in starting/stopping stopwatch); [1] valid systematic error (e.g. consistent offset in measuring L to the bob, zero error in metre rule); reject swapped/mislabelled categories. (b) [1] valid improvement addressing the random error (e.g. repeat timing and average); [1] valid improvement addressing the systematic error (e.g. always measure to bob centre, use bob diameter/2 correction, check metre rule for zero error). (c) [1] correctly identifies that 25° is too large for the small-angle (SHM) approximation to hold accurately, and that this makes the measured period too long/g too low; [1] correctly suggests repeating with a smaller amplitude (e.g. <10°).

(a) \( pV \) values: \( 150.0\times20.0=3000 \); \( 120.0\times25.0=3000 \); \( 100.0\times30.0=3000 \); \( 75.0\times40.0=3000 \); \( 60.0\times50.0=3000 \) (all in kPa cm^3). Since pV is constant (\( =3000\ \text{kPa cm}^3 \)) for every pair of readings, at constant temperature the pressure of a fixed mass of gas is inversely proportional to its volume.

(b) This relationship is Boyle's law: \( pV=\text{constant} \) (for a fixed mass of gas at constant temperature).

(c) Since \( pV=3000\ \text{kPa cm}^3 \) is constant, \( p=\dfrac{3000}{V}=\dfrac{3000}{60.0}=50.0\ \text{kPa} \).

Final answer: \( p(60.0\ \text{cm}^3)=50.0\ \text{kPa} \), consistent with Boyle's law.

(a) [1] correct calculation of pV for at least 3 of the 5 pairs; [1] all 5 values correctly shown to be equal (3000 kPa cm^3, allow rounding); [1] correct conclusion that p is inversely proportional to V at constant temperature. (b) [1] names Boyle's law and states \( pV=\text{constant} \) (both required for the mark). (c) [1] correct method (using pV=3000 and V=60.0, or direct inverse proportion from any data row); [1] \( p=50.0\ \text{kPa} \).