2022 CCEA GCSE Physics 1210 Practice Paper with Answers

(a) \( a = \frac{v-u}{t} = \frac{8.0 - 2.0}{4.0} = 1.5 \text{ m/s}^2 \)
(b) \( s = \left(\frac{u+v}{2}\right)t = \left(\frac{2.0+8.0}{2}\right)(4.0) = (5.0)(4.0) = 20 \text{ m} \)
Check by a second route: \( s = ut + \frac{1}{2}at^2 = (2.0)(4.0) + \frac{1}{2}(1.5)(4.0)^2 = 8.0 + 12.0 = 20 \text{ m} \), which agrees.
Final answer: a = 1.5 m/s², s = 20 m.

(a) Correct equation \( a=(v-u)/t \) [1]; correct answer 1.5 m/s² with unit [1].
(b) Correct equation (either \( s=\frac{(u+v)}{2}t \) or \( s=ut+\frac{1}{2}at^2 \)) [1]; correct substitution [1]; correct answer 20 m with unit [1]. Accept ecf from (a).

(a) Resultant force \( F = 12 - 4.0 = 8.0 \text{ N} \)
(b) \( a = \frac{F}{m} = \frac{8.0}{2.5} = 3.2 \text{ m/s}^2 \)
(c) \( v = u + at = 0 + (3.2)(3.0) = 9.6 \text{ m/s} \)
Check by a second route: using \( F=ma \), \( a = 8.0/2.5 = 3.2 \) m/s² confirmed; momentum change \( = Ft = 8.0 \times 3.0 = 24 \text{ kg m/s} = mv = 2.5v \Rightarrow v = 9.6 \text{ m/s} \), which agrees.
Final answer: v = 9.6 m/s.

(a) Correct resultant force 8.0 N [1].
(b) Correct equation \( a=F/m \) [1]; correct answer 3.2 m/s² [1].
(c) Correct equation \( v=u+at \) with substitution [1]; correct answer 9.6 m/s with unit [1]. Accept ecf throughout.

(a) \( p_1V_1 = p_2V_2 \)
\( (1.0 \times 10^5)(3.0 \times 10^{-4}) = p_2 (2.0 \times 10^{-4}) \)
\( p_2 = \frac{30}{2.0 \times 10^{-4}} = 1.5 \times 10^5 \text{ Pa} \)
Check: volume fell by a factor of 2/3, so pressure should rise by a factor of 3/2: \( 1.0\times10^5 \times 1.5 = 1.5\times10^5 \) Pa, which agrees.
(b) \( \rho = \frac{m}{V} \). Convert: \( m = 0.540 \text{ kg} \), \( V = 200 \text{ cm}^3 = 2.0 \times 10^{-4} \text{ m}^3 \).
\( \rho = \frac{0.540}{2.0 \times 10^{-4}} = 2700 \text{ kg/m}^3 \)
Final answer: p2 = 1.5 × 10⁵ Pa, ρ = 2700 kg/m³.

(a) Correct equation \( p_1V_1=p_2V_2 \) [1]; correct substitution [1]; correct answer 1.5 × 10⁵ Pa with unit [1].
(b) Correct unit conversion and formula \( \rho = m/V \) with substitution [1]; correct answer 2700 kg/m³ with unit [1].

(a) \( W = mgh = (250)(10)(12) = 30\,000 \text{ J} = 3.0 \times 10^4 \text{ J} \)
(b) \( P = \frac{W}{t} = \frac{30\,000}{8.0} = 3750 \text{ W} \)
Check by a second route: \( P = Fv \), where \( F = mg = 2500 \) N and average \( v = h/t = 12/8.0 = 1.5 \) m/s, giving \( P = 2500 \times 1.5 = 3750 \) W, which agrees.
Final answer: W = 3.0 × 10⁴ J, P = 3750 W.

(a) Correct equation \( W=mgh \) [1]; correct substitution [1]; correct answer 3.0 × 10⁴ J with unit [1].
(b) Correct equation \( P=W/t \) with substitution [1]; correct answer 3750 W with unit [1].

(a) Number of half-lives in 18 hours \( = 18 / 6.0 = 3 \).
\( 640 \xrightarrow{\div 2} 320 \xrightarrow{\div 2} 160 \xrightarrow{\div 2} 80 \) counts/min.
(b) One further half-life: \( 80 \xrightarrow{\div 2} 40 \) counts/min.
Check by a second route: total elapsed time 24 hours = 4 half-lives, so \( 640 / 2^4 = 640/16 = 40 \) counts/min, which agrees.
Final answer: 80 counts/min after 18 hours; 40 counts/min after 24 hours.

(a) Correct number of half-lives (3) [1]; correct halving process shown [1]; correct answer 80 counts/min [1].
(b) Correct further halving [1]; correct answer 40 counts/min [1]. Accept ecf from (a).

(a) \( v^2 = u^2 + 2as \)
\( 0 = (24)^2 + 2(-3.0)s \)
\( 0 = 576 - 6.0s \)
\( s = \frac{576}{6.0} = 96 \text{ m} \)
(b) \( v = u + at \)
\( 0 = 24 + (-3.0)t \)
\( t = \frac{24}{3.0} = 8.0 \text{ s} \)
Check by a second route: average velocity \( = (24+0)/2 = 12 \) m/s; \( s = 12 \times 8.0 = 96 \) m, which agrees.
Final answer: s = 96 m, t = 8.0 s.

(a) Correct equation \( v^2=u^2+2as \) [1]; correct substitution/rearrangement [1]; correct answer 96 m with unit [1].
(b) Correct equation \( v=u+at \) with substitution [1]; correct answer 8.0 s with unit [1].

(a) \( \text{moment} = F \times d = 30 \times 1.5 = 45 \text{ N m} \)
(b) For balance (principle of moments): clockwise moments = anticlockwise moments.
\( F \times 2.0 = 45 \)
\( F = \frac{45}{2.0} = 22.5 \text{ N} \)
Check by a second route: since the balancing force acts further from the pivot (2.0 m vs 1.5 m), it must be smaller than 30 N; \( 22.5 \text{ N} < 30 \text{ N} \), consistent, and \( 22.5 \times 2.0 = 45 = 30 \times 1.5 \), which agrees exactly.
Final answer: moment = 45 N m, F = 22.5 N.

(a) Correct equation \( \text{moment}=F\times d \) [1]; correct answer 45 N m with unit [1].
(b) Correct application of the principle of moments (clockwise = anticlockwise) [1]; correct rearrangement [1]; correct answer 22.5 N with unit [1].

(a) Volume displaced \( = 55.0 - 40.0 = 15.0 \text{ cm}^3 \)
(b) \( \rho = \frac{m}{V} = \frac{118.5}{15.0} = 7.9 \text{ g/cm}^3 \)
(c) The calculated density (7.9 g/cm³) matches iron most closely, so the object is most likely iron.
Check by a second route: \( 7.9 \times 15.0 = 118.5 \) g, which matches the given mass exactly.
Final answer: V = 15.0 cm³, ρ = 7.9 g/cm³, the metal is iron.

(a) Correct volume 15.0 cm³ [1].
(b) Correct equation \( \rho=m/V \) with substitution [1]; correct answer 7.9 g/cm³ with unit [1].
(c) Iron identified with density comparison as reason [1]. Accept ecf from (b).

Energy supplied: \( E = Pt = 50 \times (4.0 \times 60) = 50 \times 240 = 12\,000 \text{ J} \)
Assuming all energy is transferred to the block: \( E = mc\Delta\theta \)
\( \Delta\theta = \frac{E}{mc} = \frac{12\,000}{2.0 \times 900} = \frac{12\,000}{1800} = 6.7\text{ °C (2 s.f.)} \)
Check by a second route: \( mc\Delta\theta = 2.0 \times 900 \times 6.7 = 12\,060 \text{ J} \approx 12\,000 \text{ J} \), which agrees within rounding.
Final answer: Δθ ≈ 6.7 °C.

Correct energy supplied \( E=Pt=12\,000 \) J [1]; correct equation \( E=mc\Delta\theta \) rearranged [1]; correct substitution [1]; correct answer 6.7 °C (accept 6.6–6.7 °C) with unit [1].

(a) Corrected count rate \( = 150 - 18 = 132 \text{ counts/min} \)
(b) Number of counts \( = \text{rate} \times \text{time} = 132 \times 5.0 = 660 \text{ counts} \)
Check by a second route: 132 counts/min for 5.0 min means 1.0 min gives 132, so 5.0 min gives \( 132 \times 5 = 660 \), which agrees.
Final answer: corrected rate = 132 counts/min; 660 counts in 5.0 minutes.

(a) Correct method (subtract background) [1]; correct answer 132 counts/min [1].
(b) Correct equation (counts = rate × time) [1]; correct answer 660 counts [1]. Accept ecf from (a).

Indicative content:
• Coal is a non-renewable fossil fuel; burning it releases carbon dioxide, a greenhouse gas that contributes to global warming, and sulfur dioxide, which causes acid rain.
• Coal-fired power stations can generate electricity reliably and continuously, on demand, regardless of weather conditions.
• Wind is a renewable resource and produces no greenhouse gas emissions during operation.
• Wind farms are unreliable because output depends on wind speed — no electricity is generated when there is no wind, and turbines can also be shut down in very high winds.
• Coal has ongoing fuel costs and the coal supply will eventually run out; wind has high initial set-up costs (turbines, land) but no fuel costs afterwards, so running costs are low.
• A balanced conclusion should weigh reliability against environmental impact, e.g. that a mixed-supply grid combining both is often used to balance reliability with reduced emissions.
Final answer: a well-structured comparison addressing reliability, environmental impact and cost for both methods, with a supported conclusion.

Level 3 (5–6 marks): Answer addresses reliability, environmental impact AND cost for both methods; at least 5 indicative points made; answer is well organised with correct use of specialist terms (e.g. renewable, non-renewable, greenhouse gas, emissions) and fluent, accurate written English.
Level 2 (3–4 marks): 3–4 indicative points made, covering at least two of the three themes; reasonably organised with mostly appropriate use of specialist terms.
Level 1 (1–2 marks): 1–2 indicative points made; basic, list-like response; limited use of specialist terms.
Level 0 (0 marks): No creditworthy content.

(a) Time (s) on the x-axis; velocity (m/s) on the y-axis.
(b) The velocity increases steadily with time, so the graph is a straight line passing through the origin.
(c) Gradient \( = \frac{\Delta v}{\Delta t} = \frac{8.0 - 0}{4.0 - 0} = 2.0 \text{ m/s}^2 \). The gradient of a velocity-time graph represents acceleration.
Check by a second route: using any other pair of points, e.g. between t=1.0 s and t=3.0 s: \( (6.0-2.0)/(3.0-1.0) = 4.0/2.0 = 2.0 \text{ m/s}^2 \), which agrees.
Final answer: gradient = 2.0 m/s², representing the acceleration.

(a) Both axes correctly identified with units [1].
(b) Straight line through the origin (or equivalent description of uniform acceleration) [1].
(c) Correct gradient calculation shown [1]; correct value 2.0 m/s² identified as acceleration [1].

(a) Time (s) on the x-axis; distance (m) on the y-axis.
(b) Between t = 20 s and t = 30 s the distance stays at 100 m, so the cyclist was stationary; this appears as a horizontal (flat) section of the graph, since a zero gradient means zero speed.
(c) \( \text{speed} = \frac{\Delta d}{\Delta t} = \frac{140 - 100}{40 - 30} = \frac{40}{10} = 4.0 \text{ m/s} \)
Check by a second route: distance covered (40 m) over time taken (10 s) gives the same ratio however the calculation is set up, 4.0 m/s.
Final answer: stationary between t = 20 s and 30 s; average speed 4.0 m/s.

(a) Both axes correctly identified with units [1].
(b) Correct interval 20 s to 30 s [1]; correct description (horizontal/flat line, zero gradient = zero speed) [1].
(c) Correct answer 4.0 m/s with unit [1].

(a) Load (N) on the x-axis; extension (cm) on the y-axis.
(b) From 0 N to 6.0 N, extension is proportional to load (each 2.0 N adds 1.5 cm: 1.5, 3.0, 4.5 cm). At 8.0 N, if proportionality continued the extension would be 6.0 cm, but it is actually 8.5 cm — a disproportionate jump. So the spring stops obeying Hooke's law beyond a load of 6.0 N (its elastic limit has been exceeded).
(c) Within the proportional region: \( k = \frac{F}{x} = \frac{2.0}{1.5} = 1.33 \approx 1.3 \text{ N/cm} \)
Check by a second route: using the 6.0 N point, \( k = 6.0/4.5 = 1.33 \text{ N/cm} \), which agrees.
Final answer: Hooke's law fails beyond 6.0 N; spring constant ≈ 1.3 N/cm.

(a) Both axes correctly identified with units [1].
(b) Correct load identified (6.0 N) [1]; valid reason referring to the disproportionate extension at 8.0 N [1].
(c) Correct spring constant 1.3 N/cm (accept 1.3–1.4 N/cm or equivalent in N/m) with unit [1].

(a) Time (min, or s) on the x-axis; energy gained (kJ, or J) on the y-axis.
(b) The graph is a straight line through the origin, showing that energy is transferred at a constant rate, i.e. the heater has constant power output.
(c) Gradient \( = \frac{120 \text{ kJ}}{8 \text{ min}} = \frac{120\,000 \text{ J}}{480 \text{ s}} = 250 \text{ W} \)
Check by a second route: using the point at 4 min: \( \frac{60\,000 \text{ J}}{240 \text{ s}} = 250 \text{ W} \), which agrees.
Final answer: power = 250 W.

(a) Both axes correctly identified with units [1].
(b) Straight line through origin, correctly linked to constant power/rate [1].
(c) Correct unit conversion and gradient calculation shown [1]; correct answer 250 W with unit [1].

Mass number is conserved: \( 226 = A + 4 \Rightarrow A = 222 \)
Atomic (proton) number is conserved: \( 88 = Z + 2 \Rightarrow Z = 86 \)
Check by a second route: radon has atomic number 86 on the periodic table, matching an alpha decay two places back from radium (88), which agrees.
Final answer: \( {}^{226}_{88}\text{Ra} \rightarrow {}^{222}_{86}\text{Rn} + {}^{4}_{2}\text{He} \)

Correct mass number 222 [1]; correct atomic number 86 [1]; fully correct balanced equation [1].

In beta-minus decay, a neutron in the nucleus changes into a proton and an electron (beta particle), which is emitted. Mass number is unchanged; atomic number increases by 1.
\( A = 14 \) (unchanged)
\( Z = 6 + 1 = 7 \)
Check by a second route: nitrogen's atomic number is 7 on the periodic table, one more than carbon's 6, consistent with beta-minus decay increasing the proton number by one.
Final answer: \( {}^{14}_{6}\text{C} \rightarrow {}^{14}_{7}\text{N} + {}^{0}_{-1}\text{e} \)

Correct mass number 14 [1]; correct atomic number 7 [1]; fully correct balanced equation [1].

Gamma radiation is highly penetrating, so it can pass through packaging and reach all surfaces of the equipment to kill microorganisms without the packaging needing to be opened. It also does not make the sterilised equipment radioactive, so it is safe to handle afterwards, unlike using a source that leaves radioactive contamination.
Final answer: highly penetrating (passes through packaging); does not make the equipment radioactive.

Any two valid properties, e.g. highly penetrating / passes through packaging [1]; does not make equipment radioactive / no residual contamination [1]. Accept other valid relevant properties.

Newton's first law: an object remains at rest, or moves with constant velocity in a straight line, unless a resultant (unbalanced) force acts on it.
Newton's second law: the acceleration of an object is proportional to the resultant force acting on it and inversely proportional to its mass, \( F = ma \).
Newton's third law: when two objects interact, they exert equal and opposite forces on each other (for every action there is an equal and opposite reaction).
Final answer: as stated above for the three laws.

First law correctly stated (constant velocity/rest unless resultant force) [1]; second law correctly stated (F = ma or equivalent) [1]; third law correctly stated (equal and opposite forces/action-reaction pair) [1].

By Newton's third law, when two objects interact they exert equal and opposite forces on each other. So the force the lorry exerts on the car is equal in magnitude but opposite in direction to the force the car exerts on the lorry, even though the car and lorry may have very different masses and experience different accelerations as a result.
Final answer: the forces are equal in size and opposite in direction (an action-reaction pair).

Correct comparison: forces equal in size and opposite in direction [1]; correct reference to Newton's third law / action-reaction pair as the reason [1].

Terminal velocity is the maximum, constant velocity reached by an object (e.g. falling through air or another fluid) when the resistive/drag force acting on it has increased to become equal in size to its weight, so the resultant force on the object is zero and it no longer accelerates.
Final answer: the constant maximum velocity reached when weight equals drag (resultant force = zero).

Constant/maximum velocity identified [1]; correct condition given (resultant force is zero / weight equals drag or air resistance) [1].

Velocity is the rate of change of an object's displacement (its speed in a given direction), measured in m/s. Acceleration is the rate of change of an object's velocity, measured in m/s². An object can have a constant, non-zero velocity with zero acceleration, or can accelerate while at any velocity including zero, showing they are distinct quantities.
Final answer: velocity measures rate of change of displacement; acceleration measures rate of change of velocity, so they are different physical quantities with different units.

Correct definition of velocity (rate of change of displacement) [1]; correct definition of acceleration (rate of change of velocity), distinguishing the two [1].

In a liquid, particles are close together and are held by strong intermolecular forces of attraction. In a gas, the particles are much further apart, moving randomly and rapidly with negligible forces of attraction between them, so they occupy a much larger volume for the same number of particles. Since density = mass / volume, and the mass of a given number of particles is unchanged while the volume is far greater in the gas state, the density of the gas is much lower.
Final answer: gas particles are far more spread out than liquid particles, so the same mass occupies a much larger volume, giving a much lower density.

Correct particle description — particles much further apart in a gas than a liquid [1]; correct link to density (same mass, larger volume → lower density) [1].

The principle of conservation of energy states that energy cannot be created or destroyed, only transferred from one store (or form) to another; the total energy in a closed system remains constant.
Example: as an object falls, energy is transferred from its gravitational potential energy store to its kinetic energy store (with some also dissipated to the thermal energy store of the surroundings due to air resistance).
Final answer: energy is conserved — it is transferred between stores, not created or destroyed; illustrated by GPE transferring to KE as an object falls.

Correct statement of conservation of energy (not created or destroyed, only transferred) [1]; valid supporting example of an energy transfer [1].

(a) On entering the glass (going from a less dense to a more dense medium), the ray slows down and bends towards the normal. Inside the block the ray travels in a straight line to the second (parallel) surface. On leaving the glass block (going from a more dense to a less dense medium), the ray speeds up and bends away from the normal, emerging parallel to the original incident ray (but laterally displaced).
(b) \( n = \frac{\sin i}{\sin r} \)
\( 1.5 = \frac{\sin 40^\circ}{\sin r} = \frac{0.643}{\sin r} \)
\( \sin r = \frac{0.643}{1.5} = 0.429 \)
\( r = \sin^{-1}(0.429) = 25.4^\circ \)
Check by a second route: since the ray goes into a denser medium, r must be smaller than i (25.4° < 40°), which agrees with bending towards the normal.
Final answer: angle of refraction ≈ 25.4°.

(a) Correct description of bending towards the normal on entry [1]; correct description of bending away from the normal on exit, emerging parallel to the incident ray [1].
(b) Correct equation \( n=\sin i/\sin r \) with substitution [1]; correct answer 25.4° (accept 25–26°) [1].

(a) The two 6.0 Ω resistors must be connected in parallel with each other (so each has the full 12 V of the battery across it), and this parallel combination is connected to the battery with the ammeter placed in the main circuit, in series with the battery, so it reads the total current drawn from the battery.
(b) Combined resistance: \( \frac{1}{R} = \frac{1}{6.0} + \frac{1}{6.0} = \frac{2}{6.0} \Rightarrow R = 3.0 \text{ Ω} \)
Total current: \( I = \frac{V}{R} = \frac{12}{3.0} = 4.0 \text{ A} \)
Check by a second route: each resistor individually carries \( I = V/R = 12/6.0 = 2.0 \) A, and since they are in parallel the total current is the sum, \( 2.0 + 2.0 = 4.0 \) A, which agrees.
Final answer: ammeter reads 4.0 A.

(a) Resistors correctly identified as connected in parallel with each other [1]; ammeter correctly placed in the main circuit in series with the battery [1].
(b) Correct combined resistance 3.0 Ω shown [1]; correct answer 4.0 A with unit [1].

(a) The iron filings would form a pattern of concentric circles centred on the wire, in planes perpendicular to the wire. If the current is increased, the magnetic field becomes stronger, so the pattern of filings becomes denser (the field lines are more closely spaced), though the shape remains concentric circles.
(b) \( F = BIl = (0.25)(3.0)(0.40) = 0.30 \text{ N} \)
Check by a second route: \( F = BIl \); multiplying step by step, \( 0.25 \times 3.0 = 0.75 \), then \( 0.75 \times 0.40 = 0.30 \) N, which agrees.
Final answer: F = 0.30 N.

(a) Concentric circles centred on the wire [1]; correct description of field becoming stronger/lines closer together with increased current [1].
(b) Correct equation \( F=BIl \) with substitution [1]; correct answer 0.30 N with unit [1].

(a) In shallower water the waves travel more slowly, so the speed decreases. Since \( v = f\lambda \) and the frequency stays constant, a decrease in speed means the wavelength must also decrease (the wavefronts get closer together).
(b) The wavefronts change direction, bending towards the normal at the boundary. This happens because the part of each wavefront that enters the shallow water first slows down while the rest is still travelling faster in the deep water, causing the wavefront (and hence the direction of travel) to bend towards the normal — this is refraction, analogous to light bending towards the normal when entering a denser (slower) medium.
Check by a second route: since \( v=f\lambda \) with f fixed, and v is smaller in shallow water, \( \lambda \) must be smaller too, confirming wavelength decreases, consistent with the wavefronts (crests) being drawn closer together and bent towards the normal.
Final answer: speed and wavelength both decrease; wavefronts bend towards the normal.

(a) Correct statement that speed decreases [1]; correct statement that wavelength decreases (with valid reasoning via \(v=f\lambda\)) [1].
(b) Correct description — wavefronts/direction bend towards the normal [1]; correct reasoning referring to the wave slowing down [1]; correctly identifies this as analogous to refraction [1].

(a) \( \sin C = \frac{1}{n} = \frac{1}{1.52} = 0.658 \)
\( C = \sin^{-1}(0.658) = 41.1^\circ \)
(b) Total internal reflection occurs — all of the light is reflected back into the glass, none is transmitted into the air.
Check by a second route: \( \sin(41.1^\circ) = 0.658 \), and \( 1/0.658 = 1.52 \), which agrees with the given refractive index.
Final answer: critical angle ≈ 41.1°; total internal reflection.

(a) Correct equation \( \sin C = 1/n \) [1]; correct substitution [1]; correct answer 41.1° (accept 41–41.2°) [1].
(b) Correct answer: total internal reflection [1].

(a) \( m = \frac{\text{image height}}{\text{object height}} = \frac{12.0}{3.0} = 4.0 \)
(b) \( m = \frac{v}{u} \Rightarrow v = m \times u = 4.0 \times 5.0 = 20 \text{ cm} \)
Check by a second route: image height / object height should equal image distance / object distance for similar triangles in the ray diagram; \( 20/5.0 = 4.0 \), matching the magnification found in (a).
Final answer: m = 4.0, v = 20 cm.

(a) Correct equation \( m=\frac{\text{image height}}{\text{object height}} \) with substitution [1]; correct answer 4.0 (no unit) [1].
(b) Correct equation \( m=v/u \) rearranged and substituted [1]; correct answer 20 cm with unit [1].

(a) \( R_{total} = 4.0 + 6.0 + 10.0 = 20.0 \text{ Ω} \) (resistors in series simply add)
(b) \( I = \frac{V}{R} = \frac{12}{20.0} = 0.60 \text{ A} \)
(c) \( V = IR = (0.60)(6.0) = 3.6 \text{ V} \)
Check by a second route: p.d.s across all three resistors should sum to 12 V: \( V_4 = 0.60\times4.0=2.4 \), \( V_6=3.6 \), \( V_{10}=0.60\times10.0=6.0 \); sum \( =2.4+3.6+6.0=12.0 \) V, which agrees exactly with the supply voltage.
Final answer: R_total = 20.0 Ω, I = 0.60 A, V(6.0Ω) = 3.6 V.

(a) Correct answer 20.0 Ω with unit [1].
(b) Correct equation \( I=V/R \) with substitution [1]; correct answer 0.60 A with unit [1].
(c) Correct answer 3.6 V with unit [1]. Accept ecf throughout.

(a) \( P = VI \Rightarrow I = \frac{P}{V} = \frac{2300}{230} = 10 \text{ A} \)
(b) \( E = Pt = 2300 \times (3.0 \times 60) = 2300 \times 180 = 414\,000 \text{ J} = 414 \text{ kJ} \)
Check by a second route: \( E = VIt = 230 \times 10 \times 180 = 414\,000 \) J, which agrees.
Final answer: I = 10 A, E = 414 kJ.

(a) Correct equation \( I=P/V \) with substitution [1]; correct answer 10 A with unit [1].
(b) Correct equation \( E=Pt \) with correct time conversion and substitution [1]; correct answer 414 kJ with unit [1].

(a) \( Q = It = 0.50 \times (6.0 \times 60) = 0.50 \times 360 = 180 \text{ C} \)
(b) \( E = QV = 180 \times 12 = 2160 \text{ J} \)
Check by a second route: \( P = VI = 12 \times 0.50 = 6.0 \) W, so \( E = Pt = 6.0 \times 360 = 2160 \) J, which agrees.
Final answer: Q = 180 C, E = 2160 J.

(a) Correct equation \( Q=It \) with correct time conversion and substitution [1]; correct answer 180 C with unit [1].
(b) Correct equation \( E=QV \) with substitution [1]; correct answer 2160 J with unit [1].

(a) \( v = f\lambda \Rightarrow \lambda = \frac{v}{f} = \frac{330}{440} = 0.75 \text{ m} \)
(b) Since \( v = f\lambda \) and v is constant, \( \lambda \) is inversely proportional to f. Doubling f halves \( \lambda \), so the ratio of the second wavelength to the first is 1:2.
Check by a second route: second wave's wavelength \( = v/f_2 = 330/880 = 0.375 \) m; ratio \( 0.375 : 0.75 = 1:2 \), which agrees.
Final answer: λ = 0.75 m; ratio 1:2.

(a) Correct equation \( v=f\lambda \) rearranged and substituted [1]; correct answer 0.75 m with unit [1].
(b) Correct ratio 1:2 [1]; correct reasoning referring to \(v=f\lambda\) with v constant [1].

The 0.80 s is the time for the pulse to travel down to the sea bed AND back up, so the time for the one-way journey is \( 0.80/2 = 0.40 \text{ s} \).
\( \text{depth} = v \times t = 1500 \times 0.40 = 600 \text{ m} \)
Check by a second route: total distance travelled by the pulse (down and back) \( = v \times 0.80 = 1500 \times 0.80 = 1200 \) m, so the depth (one way) \( = 1200/2 = 600 \) m, which agrees.
Final answer: depth = 600 m.

Correct recognition that the pulse travels the depth twice (there and back) [1]; correct halving of the time [1]; correct equation with substitution [1]; correct answer 600 m with unit [1].

(a) \( \frac{V_p}{V_s} = \frac{N_p}{N_s} \Rightarrow V_s = V_p \times \frac{N_s}{N_p} = 230 \times \frac{100}{2000} = 230 \times 0.05 = 11.5 \text{ V} \)
(b) For an ideal (100% efficient) transformer, input power = output power: \( V_p I_p = V_s I_s \)
\( I_p = \frac{V_s I_s}{V_p} = \frac{11.5 \times 4.0}{230} = \frac{46}{230} = 0.20 \text{ A} \)
Check by a second route: turns ratio \( N_p:N_s = 20:1 \); for an ideal transformer current ratio is inverse of turns ratio, so \( I_p = I_s/20 = 4.0/20 = 0.20 \) A, which agrees.
Final answer: Vs = 11.5 V, Ip = 0.20 A.

(a) Correct equation \( V_p/V_s=N_p/N_s \) with substitution [1]; correct answer 11.5 V with unit [1].
(b) Correct equation \( V_pI_p=V_sI_s \) with substitution [1]; correct answer 0.20 A with unit [1].

The satellite travels a distance equal to the circumference of its orbit, \( 2\pi r \), in one orbital period, T.
\( v = \frac{2\pi r}{T} = \frac{2\pi (7.0 \times 10^6)}{5800} = \frac{4.40 \times 10^7}{5800} = 7.6 \times 10^3 \text{ m/s} \)
Check by a second route: \( 7.6 \times 10^3 \times 5800 = 4.41 \times 10^7 \) m, and \( 2\pi \times 7.0\times10^6 = 4.40\times10^7 \) m, which agree closely (small rounding).
Final answer: v ≈ 7.6 × 10³ m/s.

Correct equation \( v=2\pi r/T \) [1]; correct substitution [1]; correct calculation of circumference [1]; correct final answer 7.6 × 10³ m/s (accept 7.5–7.6 × 10³ m/s) with unit [1].

Indicative content:
• The star forms from a nebula — a large cloud of gas (mainly hydrogen) and dust — which contracts under its own gravitational attraction.
• As it contracts and heats up, the cloud becomes a protostar.
• When the core becomes hot and dense enough, nuclear fusion of hydrogen nuclei into helium begins, releasing enormous amounts of energy; the star is now a stable main sequence star, which is the stage the Sun is in now.
• The star remains on the main sequence for a very long time while the outward pressure from fusion balances the inward pull of gravity.
• Once the hydrogen fuel in the core runs low, the core contracts and the outer layers expand and cool, and the star becomes a red giant.
• The outer layers of the red giant are eventually shed into space, forming a planetary nebula.
• The remaining core collapses under gravity to become a white dwarf, a small, extremely dense, hot remnant, which then cools and dims over a very long period of time.
Final answer: a well-structured description covering nebula → protostar → main sequence star → red giant → planetary nebula → white dwarf, with the physical processes (gravitational contraction, nuclear fusion) explained.

Level 3 (5–6 marks): At least 5 correct stages/processes identified in the correct order (nebula, protostar, main sequence, red giant, planetary nebula, white dwarf) with reference to gravitational contraction and nuclear fusion; answer well organised with fluent, accurate use of specialist terms.
Level 2 (3–4 marks): 3–4 correct stages identified, mostly in a sensible order; reasonably organised with appropriate use of some specialist terms.
Level 1 (1–2 marks): 1–2 correct stages/facts given; basic, list-like response with limited use of specialist terms.
Level 0 (0 marks): No creditworthy content.

(a) Voltage (V) on the x-axis; current (A) on the y-axis.
(b) The graph starts as a straight line through the origin at low voltage but curves, bending towards the voltage (x) axis, as the voltage increases — the gradient (which represents 1/R) decreases. This is because as more current flows, the filament gets hotter, and the resistance of the filament increases with temperature; a higher resistance means the current increases less than proportionally for each further increase in voltage.
(c) \( R = \frac{V}{I} = \frac{4.0}{1.00} = 4.0 \text{ Ω} \)
Check by a second route: at V=1.0V, R=1.0/0.40=2.5 Ω; at V=4.0V, R=4.0 Ω, confirming resistance has increased with voltage/temperature, consistent with the curved shape described.
Final answer: R = 4.0 Ω at 4.0 V.

(a) Both axes correctly identified with units [1].
(b) Correct description of curve bending towards the voltage axis / non-linear [1]; correct statement that resistance increases with temperature [1]; correct link between increasing resistance and the decreasing gradient [1].
(c) Correct answer 4.0 Ω with unit [1].

(a) sin r on the x-axis; sin i on the y-axis. The gradient of this straight line graph (through the origin) represents the refractive index, n, since \( n = \sin i / \sin r \).
(b) \( n = \frac{\sin i}{\sin r} = \frac{0.60}{0.40} = 1.5 \)
(c) Using the gradient of the full graph averages out the effect of random errors across all the data points, giving a more accurate and reliable value than relying on any single pair of readings, which may be affected by an anomalous reading.
Check by a second route: using a different pair of readings, sin i = 0.30, sin r = 0.20: \( n = 0.30/0.20 = 1.5 \), which agrees, confirming a consistent refractive index across the data.
Final answer: n = 1.5.

(a) Both axes correctly identified [1]; gradient correctly identified as the refractive index [1].
(b) Correct equation with substitution [1]; correct answer 1.5 (no unit) [1].
(c) Valid advantage referring to reducing/averaging out random error across multiple readings [1].

(a) Using \( v = f\lambda \):
Radio wave: \( v = (1.0\times10^8)(3.0) = 3.0\times10^8 \) m/s
Visible light: \( v = (6.0\times10^{14})(5.0\times10^{-7}) = 3.0\times10^8 \) m/s
X-ray: \( v = (3.0\times10^{18})(1.0\times10^{-10}) = 3.0\times10^8 \) m/s
All three give the same speed, \( 3.0 \times 10^8 \) m/s, the speed of light in a vacuum, confirming that all electromagnetic waves travel at this speed in a vacuum.
(b) Since \( v = f\lambda \) can be rearranged to \( f = v \times (1/\lambda) \), plotting f against \( 1/\lambda \) gives a straight line through the origin, with gradient equal to v, the speed of light.
Check by a second route: for the radio wave, \( f/(1/\lambda) = (1.0\times10^8)/(1/3.0) = (1.0\times10^8)(3.0) = 3.0\times10^8 \), matching the speed of light found in (a).
Final answer: all three waves travel at 3.0 × 10⁸ m/s; the graph is a straight line through the origin with gradient equal to the speed of light.

(a) Correct calculation for at least two waves showing \(v \approx 3.0\times10^8\) m/s [1]; correct conclusion that all three speeds are equal (= speed of light) [1].
(b) Correct description — straight line through the origin [1]; gradient correctly identified as the speed of light [1].

(a) Total internal reflection occurs when light travelling within a denser medium strikes the boundary with a less dense medium at an angle of incidence greater than the critical angle; instead of being refracted out, all of the light is reflected back into the denser medium.
(b) Application: optical fibres, e.g. in telecommunications cables or medical endoscopes.
Final answer: as stated above.

(a) Correct condition — angle of incidence greater than the critical angle, in a denser medium [1]; correct outcome — all light reflected, none refracted out [1].
(b) Any valid practical application (e.g. optical fibres, endoscopes) [1].

(a) A converging (convex) lens is thicker in the middle than at the edges and bends parallel rays of light inwards so that they meet (converge) at a focal point. A diverging (concave) lens is thinner in the middle than at the edges and bends parallel rays of light outwards (they appear to diverge/spread apart from a virtual focal point).
(b) e.g. a magnifying glass, a camera lens, or spectacles for correcting long-sightedness.
Final answer: as stated above.

(a) Correct description of a converging lens bringing rays together [1]; correct description of a diverging lens spreading rays apart [1].
(b) Any valid device using a converging lens [1].

The refractive index, n, of a material is defined as the ratio of the speed of light in a vacuum (or air) to the speed of light in that material, \( n = c/v \); equivalently, for light passing from air into the material, it is the ratio of the sine of the angle of incidence to the sine of the angle of refraction, \( n = \sin i / \sin r \).
Final answer: n = c/v (or equivalently n = sin i / sin r).

Correct relationship stated, either \(n=c/v\) or \(n=\sin i/\sin r\) [1]; correctly explained in words (ratio of speeds, or ratio of sines) [1].

(a) Fleming's left-hand rule.
(b) (i) Reversing the current direction reverses the direction of the force. (ii) Reversing the magnetic field direction also reverses the direction of the force.
Final answer: Fleming's left-hand rule; the force reverses direction if either the current or the field is reversed.

(a) Correct rule named (Fleming's left-hand rule) [1].
(b)(i) Correct answer — force reverses direction [1]. (ii) Correct answer — force reverses direction [1].

The strength of the magnetic field produced by a solenoid depends on: (1) the size of the current flowing through the coil (larger current gives a stronger field); (2) the number of turns on the coil (more turns gives a stronger field); (3) whether the coil has a soft-iron core inside it (a soft-iron core greatly increases the field strength compared with an air core).
Final answer: current, number of turns, and presence of a soft-iron core.

Any three valid factors from: current in the coil; number of turns; presence of a soft-iron core [1 mark each, max 3].

(a) In a transverse wave, the particles (or field disturbance) vibrate at right angles (perpendicular) to the direction in which the wave transfers energy. In a longitudinal wave, the particles vibrate parallel to (along) the direction in which the wave transfers energy.
(b) Example: sound waves (also, e.g., P-waves in an earthquake).
Final answer: as stated above; sound waves are a longitudinal wave.

(a) Correct description of transverse wave (perpendicular vibration) [1]; correct description of longitudinal wave (parallel vibration) [1].
(b) Valid example of a longitudinal wave (e.g. sound) [1].

(1) The light from distant galaxies is redshifted (shifted towards longer, redder wavelengths), showing that galaxies are moving away from us and the universe is expanding, consistent with everything having originated from a single point.
(2) Cosmic microwave background (CMB) radiation is detected coming almost uniformly from all directions in space; it is the cooled remnant of the intense radiation released shortly after the Big Bang.
Final answer: redshift of galaxies and cosmic microwave background radiation.

Any two valid pieces of evidence: redshift of light from galaxies [1]; cosmic microwave background radiation [1].

A geostationary satellite orbits the Earth directly above the equator, in the same direction as the Earth's rotation, with an orbital period of exactly 24 hours; as a result it stays above the same point on the Earth's surface at all times, as seen from the ground.
Use: e.g. satellite television broadcasting or telecommunications.
Final answer: as stated above.

Correct definition, including orbital period of 24 hours above the equator, remaining above a fixed point [1]; valid use given (e.g. satellite TV, communications) [1].

Partially fill the measuring cylinder with a known volume of water and record the level. Carefully lower the stone into the water (e.g. on a thread, tilting the cylinder to avoid splashing) so it is fully submerged, and record the new water level. The volume of the stone is the difference between the two readings.
Final answer: as stated above.

Initial water level recorded, stone fully submerged without splashing [1]; volume found as the difference between the two readings [1].

Mass: a top-pan (electronic) balance. Diameter (higher precision than a ruler): vernier calipers, or a micrometer screw gauge.
Final answer: top-pan balance for mass; vernier calipers/micrometer for diameter.

Correct apparatus for mass (top-pan/electronic balance) [1]; correct apparatus for diameter (vernier calipers or micrometer) [1].

(a) The ammeter measures the size of the current flowing through the coil, so that it can be recorded and kept constant (or set to specific values) during the investigation.
(b) The variable resistor allows the current in the circuit to be adjusted/controlled, so that its effect on the strength of the electromagnet can be investigated, or so the current can be kept constant when another variable (e.g. number of turns) is changed.
Final answer: as stated above.

(a) Correct purpose — to measure/monitor the current [1]. (b) Correct purpose — to vary/control the current [1].

Set up the circuit and use the variable resistor to set the current to a fixed value, checked using the ammeter. For a given number of turns of wire on the nail, bring the electromagnet close to a pile of small steel paperclips and count the maximum number it can pick up and hold. Reset and repeat this test for different numbers of turns of wire (e.g. 10, 20, 30, 40 turns), keeping the current and all other variables (e.g. type of core, distance to the paperclips) the same each time, then compare the maximum number of paperclips picked up for each number of turns.
Final answer: as stated above.

Correct method described — counting the maximum number of paperclips picked up [1]; current kept constant using the ammeter/variable resistor [1]; test repeated systematically for different numbers of turns [1].

(1) View the water level at eye level, level with the bottom of the meniscus, to avoid parallax error. (2) Make sure the stone is fully submerged and does not touch the sides or bottom of the cylinder, which could give an inaccurate reading. (3) Lower the stone into the water gently and slowly (e.g. on a thread) to avoid splashing water out of the cylinder, which would cause the volume reading to be too low.
Final answer: as stated above.

Any three valid precautions, e.g. reading at eye level to avoid parallax error; stone fully submerged, not touching sides/bottom; lowered gently to avoid splashing [1 mark each, max 3].

To make the investigation a fair test of the effect of the number of turns, the following should be kept constant: (1) the current flowing through the coil; (2) the type and size (e.g. length, material) of the iron core (nail) used; (3) the distance between the electromagnet and the paperclips (or other object) being used to test its strength.
Final answer: as stated above.

Any three valid controlled variables, e.g. current; type/size of core; distance to test object [1 mark each, max 3].

\( \rho = m/V \)
Sample A: \( 53.0/20.0 = 2.65 \text{ g/cm}^3 \)
Sample B: \( 82.5/30.0 = 2.75 \text{ g/cm}^3 \)
Sample C: \( 106.8/40.0 = 2.67 \text{ g/cm}^3 \)
(b) The three density values are all close to one another (within experimental variation), even though the samples are different sizes. This shows that density is an intrinsic property of the material and does not depend on the amount (mass or volume) of the sample.
Check by a second route: \( 2.65 \times 20.0 = 53.0 \) g and \( 2.75\times30.0=82.5 \) g and \( 2.67\times40.0=106.8 \) g, each matching the given mass, confirming the divisions are correct.
Final answer: A = 2.65 g/cm³, B = 2.75 g/cm³, C = 2.67 g/cm³; density is consistent regardless of sample size.

(a) At least two of the three values correct [1]; all three values correct (2.65, 2.75, 2.67 g/cm³) [1].
(b) Valid comment that the densities are similar/consistent across different sample sizes, showing density is independent of sample size [1]. Accept ecf from (a).

(a) Average \( = \frac{2.65 + 2.75 + 2.67}{3} = \frac{8.07}{3} = 2.69 \text{ g/cm}^3 \)
(b) \( \text{percentage error} = \frac{|\text{measured} - \text{accepted}|}{\text{accepted}} \times 100\% = \frac{|2.69 - 2.70|}{2.70} \times 100\% = \frac{0.01}{2.70} \times 100\% \approx 0.37\% \)
Check by a second route: \( 2.70 \times 0.0037 = 0.01 \), and \( 2.70 - 0.01 = 2.69 \), which agrees with the measured average.
Final answer: average density = 2.69 g/cm³; percentage error ≈ 0.37%.

(a) Correct average 2.69 g/cm³ [1].
(b) Correct equation for percentage error [1]; correct answer ≈ 0.37% (accept 0.3–0.4%) [1]. Accept ecf from (a).

(a) As the number of turns doubles (10 → 20), the number of paperclips picked up also doubles (3 → 6); as it triples (10 → 30), the paperclips triple (3 → 9). The number of paperclips picked up is directly proportional to the number of turns.
(b) Using the ratio 3 paperclips per 10 turns: at 40 turns, \( \text{paperclips} = 3 \times \frac{40}{10} = 3 \times 4 = 12 \)
Check by a second route: the pattern increases by 3 paperclips for every 10 extra turns (3, 6, 9, ...), so the next value after 9 (at 30 turns) is \( 9 + 3 = 12 \) at 40 turns, which agrees.
Final answer: directly proportional; 12 paperclips at 40 turns.

(a) Correct relationship identified — directly proportional [1].
(b) Correct method shown (using the proportional pattern) [1]; correct answer 12 [1].

(a) As the current increases, the number of paperclips picked up increases in direct proportion (doubling the current from 0.20 A to 0.40 A doubles the paperclips from 2 to 4, and so on), so the number of paperclips picked up is directly proportional to the current.
(b) The method would work because steel paperclips are magnetic (attracted to the magnetic field produced by the electromagnet) and so are picked up, while plastic-coated (non-metallic) items are not magnetic and are not attracted, so they are left behind. Limitation: this simple electromagnet can only pick up a limited number/mass of paperclips at a time (limited strength), so it is not practical for sorting large quantities quickly without being scaled up considerably.
Final answer: directly proportional; works because only magnetic (steel) items are attracted; limited by the small maximum load it can lift.

(a) Correct relationship identified — directly proportional [1].
(b) Valid reason referring to steel being magnetic/attracted while plastic is not [1]; valid limitation (e.g. limited pick-up capacity/strength, or requires continuous power) [1].

The graph would be a straight line, passing (approximately, allowing for small experimental scatter) through the origin. This shape shows that mass is directly proportional to volume for samples of the same material — doubling the volume doubles the mass, and so on.
Final answer: straight line through the origin; mass is directly proportional to volume.

Correct shape identified — straight line through the origin [1]; correct interpretation — mass directly proportional to volume [1].

Since \( \rho = m/V \), and the gradient of a mass (y-axis) versus volume (x-axis) graph is also \( \Delta m / \Delta V \), the gradient of this graph is equal to the density of the material. Using the average density found earlier, the gradient \( = 2.69 \text{ g/cm}^3 \).
Final answer: gradient = density = 2.69 g/cm³.

Correct statement that the gradient equals the density [1]; correct value 2.69 g/cm³ with unit [1]. Accept ecf from earlier average.

(a) The graph is a straight line through the origin (since paperclips picked up is directly proportional to the number of turns).
(b) Gradient \( = \frac{\Delta y}{\Delta x} = \frac{12 - 3}{40 - 10} = \frac{9}{30} = 0.30 \text{ paperclips per turn} \)
Check by a second route: using the origin and the point at 20 turns: \( (6-0)/(20-0) = 0.30 \), which agrees.
Final answer: straight line through the origin; gradient = 0.30 paperclips per turn.

(a) Correct shape — straight line through the origin [1].
(b) Correct gradient calculation shown [1]; correct answer 0.30 paperclips/turn with unit [1].

(a) The graph is a straight line through the origin (paperclips picked up is directly proportional to current).
(b) Gradient \( = \frac{\Delta y}{\Delta x} = \frac{8 - 2}{0.80 - 0.20} = \frac{6}{0.60} = 10 \text{ paperclips per amp} \)
This gradient represents how sensitively the strength of the electromagnet (measured by paperclips picked up) responds to a change in current — a higher gradient means the magnet's strength increases more rapidly per amp of current.
Check by a second route: using the origin and the point at 0.40 A: \( (4-0)/(0.40-0)=10 \), which agrees.
Final answer: straight line through the origin; gradient = 10 paperclips/A, representing the increase in magnet strength per unit current.

(a) Correct shape — straight line through the origin [1].
(b) Correct gradient calculation shown, 10 paperclips/A with unit [1]; valid statement of what the gradient represents [1].

Indicative content:
• Equipment: a length of resistance wire (e.g. constantan) taped along a metre rule, a low-voltage power supply, an ammeter, a voltmeter, crocodile clips, and connecting leads.
• Method: connect one crocodile clip at the start of the wire (0 cm) and the other at a chosen length (e.g. 10 cm), with the ammeter in series and the voltmeter connected across this length of wire; switch on and record the ammeter and voltmeter readings, then calculate resistance using \( R=V/I \).
• Move the second crocodile clip to increase the length in equal steps (e.g. every 10 cm up to 100 cm), repeating the current and voltage measurement and resistance calculation at each length.
• Controlled variables: use the same wire throughout (same material and the same cross-sectional area/diameter), and keep the current low, switching off between readings, to avoid the wire heating up and its resistance changing.
• Reliability: repeat each measurement (e.g. three times) and calculate a mean resistance at each length; plot a graph of resistance against length to check the results follow a consistent trend and to identify any anomalous readings.
Final answer: a clear, ordered plan covering equipment, method, controlled variables and reliability measures as above.

Level 3 (5–6 marks): Plan covers appropriate equipment, a clear method for varying length and measuring resistance (via V and I), at least two correctly controlled variables, and a valid reliability measure (repeats/mean or graphing to spot anomalies); answer well organised with fluent, accurate use of specialist terms.
Level 2 (3–4 marks): Most of the above included but with some gaps (e.g. missing a controlled variable or reliability measure); reasonably organised with appropriate use of some specialist terms.
Level 1 (1–2 marks): Only a basic outline given (e.g. equipment list only, or method without controls); limited use of specialist terms.
Level 0 (0 marks): No creditworthy content.

(a) The result at 30 cm (7.5 Ω) is anomalous — the other results increase by about 2.0 Ω for every 10 cm (2.0, 4.1, [6.0 expected], 8.0, 9.9, 12.0), but the 30 cm reading (7.5 Ω) does not fit this pattern. A possible cause is a poor/loose electrical connection at the crocodile clip for that particular reading, adding extra contact resistance.
(b) A straight line of best fit passing through (or very close to) the origin.
(c) Using the points (10, 2.0) and (60, 12.0): \( \text{gradient} = \frac{12.0 - 2.0}{60 - 10} = \frac{10.0}{50} = 0.20 \text{ Ω/cm} \)
(d) \( R = \text{gradient} \times L = 0.20 \times 100 = 20 \text{ Ω} \)
Check by a second route: using the points (20, 4.1) and (50, 9.9): gradient \( = (9.9-4.1)/(50-20) = 5.8/30 = 0.193 \approx 0.19\text{-}0.20 \text{ Ω/cm} \), consistent with the value found in (c).
Final answer: anomaly at 30 cm; gradient = 0.20 Ω/cm; predicted resistance at 100 cm = 20 Ω.

(a) Correct anomaly identified (30 cm, 7.5 Ω) [1]; valid experimental cause suggested [1].
(b) Correct description — straight line through the origin [1].
(c) Correct method using two non-anomalous points [1]; correct answer 0.20 Ω/cm (accept 0.19–0.20) with unit [1].
(d) Correct answer 20 Ω with unit, consistent with gradient used [1].

(a) The line of best fit should be a straight line passing through the origin. This is because when the current is zero, there is no force on the wire (from \( F=BIl \), \( I=0 \Rightarrow F=0 \)).
(b) Using the points (0.50, 0.10) and (2.50, 0.51): \( \text{gradient} = \frac{0.51 - 0.10}{2.50 - 0.50} = \frac{0.41}{2.00} = 0.205 \approx 0.20 \text{ N/A} \)
(c) Since \( F = BIl \), the gradient of a force-current graph equals \( Bl \).
\( Bl = 0.20 \Rightarrow B = \frac{0.20}{l} = \frac{0.20}{0.50} = 0.40 \text{ T} \)
Check by a second route: using B = 0.40 T and l = 0.50 m, at I = 2.00 A, predicted \( F = 0.40\times2.00\times0.50 = 0.40 \) N, close to the measured 0.42 N, which agrees within experimental variation.
Final answer: gradient ≈ 0.20 N/A; B ≈ 0.40 T.

(a) Correct description — straight line through the origin [1]; correct physical reason (zero current gives zero force) [1].
(b) Correct method shown [1]; correct answer ≈ 0.20 N/A (accept 0.19–0.21) with unit [1].
(c) Correct rearrangement of \(F=BIl\) [1]; correct answer ≈ 0.40 T with unit [1]. Accept ecf from (b).

(a) For a simple pendulum, \( T = 2\pi\sqrt{L/g} \), so T is proportional to the square root of L, not directly proportional to L — this non-linear relationship produces a curve. Squaring both sides gives \( T^2 = \frac{4\pi^2}{g}L \), so T² is directly proportional to L, giving a straight line through the origin.
(b) Using the points (0.20, 0.81) and (1.00, 4.04): \( \text{gradient} = \frac{4.04 - 0.81}{1.00 - 0.20} = \frac{3.23}{0.80} = 4.04 \text{ s}^2\text{/m} \)
(c) \( \text{gradient} = \frac{4\pi^2}{g} \Rightarrow g = \frac{4\pi^2}{\text{gradient}} = \frac{39.48}{4.04} = 9.78 \approx 9.8 \text{ m/s}^2 \)
Check by a second route: using the points (0.40, 1.61) and (0.80, 3.24): gradient \( = (3.24-1.61)/(0.80-0.40) = 1.63/0.40 = 4.08 \text{ s}^2\text{/m} \), close to the value found in (b), and \( g = 39.48/4.08 = 9.68 \text{ m/s}^2 \), consistent (both close to the accepted value of 9.8 m/s²).
Final answer: gradient ≈ 4.0 s²/m; g ≈ 9.8 m/s².

(a) Correct reason for the curve (T proportional to √L) [1]; correct reason for the straight line (T² proportional to L) [1].
(b) Correct method shown [1]; correct answer in the range 4.0–4.1 s²/m with unit [1].
(c) Correct rearrangement of the gradient equation [1]; correct answer in the range 9.6–10.0 m/s² with unit [1]. Accept ecf from (b).

(a) The ammeter must be connected in series with R (so that exactly the same current flows through both the ammeter and R). The voltmeter must be connected in parallel with (across) R (so that it reads the potential difference across R directly).
(b) \( R = \frac{V}{I} \)
(c) The ammeter should have a very low resistance so that it does not itself add significantly to the total resistance of the circuit or reduce the current being measured. The voltmeter should have a very high resistance so that only a negligible current flows through it (rather than through R), meaning it does not significantly affect the current flowing through R.
Final answer: as stated above.

(a) Ammeter correctly placed in series with R [1]; voltmeter correctly placed in parallel with R [1].
(b) Correct equation \( R=V/I \) [1].
(c) Correct explanation referring to minimising the meters' effect on the circuit (low ammeter resistance, high voltmeter resistance) [1].

A variable resistor (rheostat) should be used. It is connected in series with the lamp (and the ammeter) in the main circuit. Adjusting the variable resistor changes the total resistance of the circuit, which changes the current flowing, and therefore changes the potential difference across the lamp, allowing a range of V and I values to be recorded.
Final answer: variable resistor, connected in series with the lamp.

Correct component named (variable resistor/rheostat) [1]; correctly described as connected in series with the lamp [1]; correct reasoning — adjusting it changes the current and hence the p.d. across the lamp [1].

Hazard: risk of electric shock from exposed or damaged wiring/connections. Precaution: check that all wires and connections are properly insulated before switching on, and switch off the circuit between readings rather than leaving it connected.
Further precaution (heating over time): the coil of wire can become hot due to resistive heating if current flows through it for an extended period, which is a burn/fire risk; the current should therefore be switched off between readings and the coil not left connected for longer than necessary.
Final answer: as stated above.

Valid electrical hazard identified [1]; valid precaution for that hazard [1]; valid further precaution relevant to prolonged current/heating (e.g. switching off between readings) [1].

(a) Group 4's result (3.15 g/cm³) is anomalous, as it is far from the other three results, which cluster closely around 2.69–2.71 g/cm³.
(b) A possible reason: Group 4 may have misread the volume scale on the measuring cylinder (parallax error), not fully submerged the object before reading the water level (giving a volume that is too small and so a density that is too high), or made an arithmetic error when calculating density.
(c) The anomalous result should be excluded from the class average (or re-measured to check if it is genuine before including it), because including such a clear outlier would skew the average away from the true value represented by the consistent results of the other three groups.
Final answer: Group 4 is anomalous; likely due to a measurement/reading error; it should be excluded (or re-checked) when calculating the class average.

(a) Correct group identified (Group 4) [1].
(b) Valid experimental reason for the anomaly [1].
(c) Correct judgement (exclude/re-check) with valid justification [1].

(a) The value 7.2 Ω is anomalous — it is far higher than the other four readings, which are all close to 4.8–5.0 Ω.
(b) Mean \( = \frac{4.8 + 4.9 + 5.0 + 4.9}{4} = \frac{19.6}{4} = 4.9 \text{ Ω} \)
Check by a second route: the four remaining values are all within 0.1 Ω of 4.9 Ω (4.8, 4.9, 5.0, 4.9), so a mean of 4.9 Ω is consistent with the spread of the data.
Final answer: anomaly = 7.2 Ω; mean (excluding anomaly) = 4.9 Ω.

(a) Correct anomaly identified (7.2 Ω) [1].
(b) Correct method (sum of 4 values divided by 4) [1]; correct answer 4.9 Ω with unit [1].

A random error causes readings to scatter unpredictably above and below the true value each time a measurement is repeated (e.g. due to human reaction time when using a stopwatch); its effect can be reduced by repeating readings and taking a mean. A systematic error causes all readings to be shifted consistently in the same direction away from the true value (e.g. due to a zero error on an instrument, or an instrument that is not calibrated correctly); it is not reduced by repeating readings and averaging.
A systematic error can be identified, for example, from a graph that is expected to pass through the origin but instead has a non-zero y-intercept — this constant offset indicates a systematic (zero) error in the measurements.
Final answer: as stated above.

Correct description of random error (unpredictable scatter, reduced by averaging) [1]; correct description of systematic error (consistent shift, not reduced by averaging) [1]; valid method of identifying a systematic error (e.g. non-zero intercept on a graph expected to pass through the origin) [1].

(a) Within the range of data actually collected (10 to 40 turns), the results do support the conclusion that strength increases with the number of turns — but the data does not extend beyond 40 turns, so it cannot prove the pattern continues indefinitely.
(b) It would not be valid to extrapolate this trend to very large numbers of turns without further testing, because the relationship may not continue to hold outside the range that was actually tested. For example, the iron core could become magnetically saturated at high numbers of turns, so it cannot be magnetised any further, and/or a much longer coil of wire has a greater resistance, which (for a fixed supply voltage) reduces the current flowing and so could reduce, not increase, the strength of the magnet.
Final answer: the data supports the trend only within the tested range (10–40 turns); extrapolating to 1000 turns is invalid due to possible core saturation and/or increased coil resistance reducing the current.

(a) Correct judgement — data supports the trend within the tested range only [1].
(b) Correctly identifies that extrapolation beyond the tested range is not valid [1]; valid physical reasoning given (e.g. core saturation, or increased resistance reducing current) [1].

(a) Mean \( = \frac{15.0 + 15.4}{2} = \frac{30.4}{2} = 15.2 \text{ cm}^3 \)
(b) \( \text{percentage difference} = \frac{|15.4 - 15.0|}{15.2} \times 100\% = \frac{0.4}{15.2} \times 100\% \approx 2.6\% \)
Check by a second route: half the difference between the readings is \( 0.2 \) cm³, and \( 0.2/15.2 \times 100 \approx 1.3\% \) on each side of the mean, giving a total spread of about 2.6% between the two readings, which agrees.
Final answer: mean = 15.2 cm³; percentage difference ≈ 2.6%.

(a) Correct mean 15.2 cm³ [1].
(b) Correct equation for percentage difference [1]; correct answer ≈ 2.6% (accept 2.5–2.7%) [1].

(a) Resistance is directly proportional to length (R ∝ L).
(b) The original 100 cm wire has resistance \( R = 0.20 \times 100 = 20 \text{ Ω} \). Cutting it exactly in half gives two 50 cm lengths; since resistance is directly proportional to length, each half has resistance \( R = 0.20 \times 50 = 10 \text{ Ω} \), exactly half the resistance of the original wire.
(c) e.g. repeat each length/resistance measurement several times and take a mean to reduce the effect of random error, or take more (closer-spaced) length values to give a more reliable line of best fit.
Check by a second route: since resistance is proportional to length, halving the length must exactly halve the resistance: \( 20/2 = 10 \text{ Ω} \), which agrees.
Final answer: R ∝ L; resistance of each half = 10 Ω.

(a) Correct relationship stated — directly proportional [1].
(b) Correct answer 10 Ω [1]; correct reasoning referring to R ∝ L (halving length halves resistance) [1].
(c) Valid method to improve reliability (e.g. repeats and mean, or more data points) [1].

Hazard: water from the measuring cylinder could spill onto the bench and come into contact with the electrical circuit (battery, wires, ammeter), creating a risk of electric shock or a short-circuit.
Management: keep the water-based density apparatus and the electrical electromagnet circuit set up in separate areas of the bench, away from each other, and mop up any spilled water immediately before continuing with the electrical part of the practical.
Final answer: as stated above.

Valid hazard identified (water near electrical apparatus → shock/short-circuit risk) [1]; valid precaution described, specific and practical (e.g. keep apparatus separated, mop up spills) [2].

(a) \( \text{percentage uncertainty} = \frac{\text{absolute uncertainty}}{\text{measured value}} \times 100\% = \frac{0.1}{85.0} \times 100\% \approx 0.12\% \)
(b) \( \text{percentage uncertainty} = \frac{0.01}{0.42} \times 100\% \approx 2.4\% \)
Check: the diameter measurement (much smaller value, same order of absolute uncertainty relative to resolution) has a much larger percentage uncertainty than the length measurement, which makes sense since percentage uncertainty is larger for smaller measured quantities with a similar absolute uncertainty.
Final answer: length ≈ 0.12%; diameter ≈ 2.4%.

(a) Correct equation with substitution [1]; correct answer ≈ 0.12% [1].
(b) Correct equation with substitution [1]; correct answer ≈ 2.4% [1].

(a) \( E_p = mgh = 0.20 \times 10 \times 4.5 = 9.0 \text{ J} \)
(b) \( \text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} \times 100\% = \frac{9.0}{15.0} \times 100\% = 60\% \)
Check by a second route: \( 15.0 \times 0.60 = 9.0 \) J, which agrees with the useful output calculated in (a).
Final answer: useful output = 9.0 J; efficiency = 60%.

(a) Correct equation \( E_p=mgh \) with substitution [1]; correct answer 9.0 J with unit [1].
(b) Correct equation for efficiency with substitution [1]; correct answer 60% [1]. Accept ecf from (a).

(a) \( V = 5.00 \times 2.00 \times 1.50 = 15.0 \text{ cm}^3 \)
(b) \( \rho = \frac{m}{V} = \frac{41.0}{15.0} = 2.73 \text{ g/cm}^3 \)
(c) \( \text{percentage error} = \frac{|2.73 - 2.70|}{2.70} \times 100\% = \frac{0.03}{2.70} \times 100\% \approx 1.1\% \)
Check by a second route: \( 2.70 \times 1.011 \approx 2.73 \), which agrees with the calculated density.
Final answer: V = 15.0 cm³, ρ = 2.73 g/cm³, percentage error ≈ 1.1%.

(a) Correct answer 15.0 cm³ with unit [1].
(b) Correct equation \( \rho=m/V \) with substitution [1]; correct answer 2.73 g/cm³ with unit [1].
(c) Correct answer ≈ 1.1% (accept 1.0–1.2%) [1]. Accept ecf throughout.

(a) \( \text{paperclips} = \text{gradient} \times \text{turns} = 0.30 \times 70 = 21 \)
(b) \( \text{mass} = \text{number of paperclips} \times \text{mass per paperclip} = 21 \times 0.50 = 10.5 \text{ g} \)
Check by a second route: at 40 turns the model predicts \( 0.30 \times 40 = 12 \) paperclips (matching the earlier data); scaling up to 70 turns by the same ratio, \( 12 \times (70/40) = 21 \), which agrees.
Final answer: 21 paperclips; total mass = 10.5 g.

(a) Correct method shown [1]; correct answer 21 [1].
(b) Correct method shown [1]; correct answer 10.5 g with unit [1]. Accept ecf from (a).

(a) \( R = \text{gradient} \times L = 0.20 \times 60 = 12 \text{ Ω} \)
(b) \( I = \frac{V}{R} = \frac{6.0}{12} = 0.50 \text{ A} \)
(c) \( P = VI = 6.0 \times 0.50 = 3.0 \text{ W} \)
Check by a second route: \( P = I^2R = (0.50)^2 \times 12 = 0.25 \times 12 = 3.0 \text{ W} \), which agrees.
Final answer: R = 12 Ω, I = 0.50 A, P = 3.0 W.

(a) Correct answer 12 Ω with unit [1].
(b) Correct equation \( I=V/R \) with substitution [1]; correct answer 0.50 A with unit [1].
(c) Correct answer 3.0 W with unit [1]. Accept ecf throughout.

(a) \( \text{percentage error} = \frac{|9.78 - 9.81|}{9.81} \times 100\% = \frac{0.03}{9.81} \times 100\% \approx 0.31\% \)
(b) Yes — a percentage error this small (well under 1%) suggests the experimental method was reliable, with only small random and/or systematic errors affecting the result.
Check by a second route: \( 9.81 \times (1 - 0.0031) = 9.81 \times 0.9969 \approx 9.78 \), which agrees with the measured value.
Final answer: percentage error ≈ 0.31%; the method was reliable.

(a) Correct equation with substitution [1]; correct answer ≈ 0.31% [1].
(b) Valid, justified comment on reliability based on the small percentage error [1].