2025 OCR GCSE Gateway Science - Physics A - J249 模拟试题及答案详解

Density = mass / volume. Mass is measured with a balance. The volume of an irregular solid is found from the rise in water level in a measuring cylinder. Option A is the only pair that gives both quantities.

A (1 mark). ALLOW any answer that clearly selects A. B, C and D do not give both mass and volume (AO2).

An ammeter measures the current through a component so must be in series with it. A voltmeter measures the potential difference across a component so must be in parallel with it.

A (1 mark). Ammeter in series and voltmeter in parallel (AO1).

\( \frac{V_p}{V_s} = \frac{N_p}{N_s} \) so \( N_s = 500 \times \frac{23}{230} = 50 \) turns.

C (1 mark). Correct use of the turns-ratio equation gives 50. Common errors: 5000 from inverting the ratio (AO2).

A step-up transformer raises the potential difference. For the same power (P = VI) the current is smaller, so less energy is transferred as heat in the transmission cables.

C (1 mark). DO NOT ALLOW the idea that a transformer increases power (AO1).

Useful output = 240 − 180 = 60 J. Efficiency = 60 / 240 = 0.25 = 25%.

A (1 mark). 75% is the common error from using the wasted energy (AO2).

\( \rho = \frac{m}{V} = \frac{0.60}{0.00025} = 2400 \) kg/m\(^3\).

C (1 mark). Correct substitution and evaluation (AO2).

GPE = m g h = 3.0 × 10 × 4.0 = 120 J.

C (1 mark). Correct use of mgh (AO2).

Light from galaxies moving away is red-shifted: its wavelength is increased. Galaxies further away show more red-shift, which is evidence that space is expanding.

A (1 mark). B describes blue-shift, which would indicate approach. DO NOT ALLOW changes in speed of light (AO1).

Density = mass / volume. The mass is measured with a balance. The volume of an irregular solid is found from the rise in water level in a measuring cylinder. So the answer is C.

C (1 mark). Distractors: A cannot give mass or volume; B measures temperature and force; D gives no mass. ALLOW any equivalent apparatus choice only if the answer letter is C.

\( V_p / V_s = N_p / N_s \), so \( N_s = N_p \times V_s / V_p = 500 \times 23 / 230 = 50 \).

A (1 mark). Correct method: substitution into the turns-ratio equation and evaluation to 50. DO NOT ALLOW other letters.

Power = potential difference x current. For the same power, a higher potential difference means a smaller current. A smaller current causes less heating of the cables (\( P = I^2R \)), so less energy is wasted.

D (1 mark). A, B and C are incorrect: step-up transformers do not change cable resistance, do not convert a.c. to d.c., and lower rather than raise the current.

Efficiency = useful output / total input = 600 / 800 = 0.75 = 75%.

C (1 mark). ALLOW 0.75 as the working. A is the wasted fraction; D is the inverted ratio.

Density = mass / volume = 540 / 200 = 2.7 g/cm\(^3\).

B (1 mark). A results from dividing volume by mass; C and D are arithmetic errors.

Total resistance = 4.0 + 8.0 = 12 \(\Omega\). \( I = V / R = 6.0 / 12 = 0.50 \) A.

A (1 mark). B uses only the 4.0 \(\Omega\) resistor (6.0 / 4.0); C and D are not obtainable from a correct method (ALLOW no other letters).

\( E_k = \frac{1}{2} m v^2 = 0.5 \times 3.0 \times 4.0^2 = 24 \) J.

B (1 mark). A uses mv; C omits the half; D is a further error.

(a) \( \frac{V_p}{V_s} = \frac{N_p}{N_s} \), so \( N_s = \frac{N_p \times V_s}{V_p} = \frac{2300 \times 12}{230} = 120 \) turns.

(b) 100% efficient, so power in = power out: \( V_p I_p = V_s I_s \). \( I_p = \frac{12 \times 1.5}{230} = \frac{18}{230} = 0.0783 \) A = 0.078 A (2 s.f.).

(c) A d.c. supply gives a steady current, so the magnetic field in the iron core is constant. A changing magnetic field (changing flux linkage) is needed to induce a potential difference in the secondary coil, so none is induced.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| (a) | Correct substitution into \( V_p/V_s = N_p/N_s \), e.g. \( 230/12 = 2300/N_s \) (1); 120 (turns) (1) | 2 | AO2 | ALLOW 1 mark for rearrangement or substitution. Correct answer with no working scores 2. ALLOW 120.0. |
| (b) | \( V_p I_p = V_s I_s \) with correct substitution, e.g. \( 230 \times I_p = 12 \times 1.5 \) (1); 0.078 (A) (1) | 2 | AO2 | ALLOW 0.0783 or 0.07826 for the evaluation mark if not rounded. DO NOT ALLOW 0.08 as the final answer given the 2 s.f. instruction (1 mark max for the method). ECF the power calculation only if the substitution is correct. |
| (c) | d.c. gives a constant/steady current (in the primary) so the magnetic field/flux (in the core) does not change (1) AND so no potential difference/current is induced in the secondary coil (AW) | 1 | AO1 | Both ideas needed for the mark: constant field/current and no induced p.d. ALLOW no changing magnetic field. IGNORE 'battery runs out'. DO NOT ALLOW 'd.c. cannot flow through the coil'. |
| Total | | 5 | | |

(a) A d.c. supply gives a constant magnetic field, so no p.d. is induced in the secondary coil.
(b) \( V_s = V_p \times N_s / N_p = 230 \times 150 / 2300 = 15 \) V.
(c) \( I_p = V_s I_s / V_p = (15 \times 2.0)/230 = 0.130 \) A (0.13 A).
(d) The transformer is not 100% efficient: energy is wasted as heat in the coils and core, so more input power is needed than the output power.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| (a) | d.c. produces a constant/unchanging magnetic field (in the core) so no p.d./current is induced in the secondary coil (AW) | 1 | AO1 | ALLOW no changing flux. DO NOT ALLOW 'no current flows' alone. |
| (b) | rearrangement/substitution: \( 230 \times 150 / 2300 \) (1); 15 (V) (1) | 2 | AO2 | Correct answer scores 2. ALLOW 1 mark for 1.5 or 1500 (power of ten error). |
| (c) | substitution: \( 230 \times I_p = 15 \times 2.0 \) (1); 0.13 (A) (1) | 2 | AO2 | ALLOW 0.130 / 0.1304. ECF from (b) for the final mark. |
| (d) | energy/power is wasted/dissipated as thermal energy (in the coils/core) so input power is greater than output power (AW) | 1 | AO3 | ALLOW resistance of coils / eddy currents. IGNORE 'friction'. |
| Total | | 6 | | |

(a) \( E = P t = 60 \times 300 = 18\,000 \) J.
(b) Change in temperature = 60 − 20 = 40 °C. \( c = \Delta E / (m \Delta\theta) = 18\,000 / (0.50 \times 40) = 900 \) J/kg°C.
(c) Wrap the block in insulation so less energy is lost to the surroundings, making the energy supplied closer to the energy gained by the block.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| (a) | \( 60 \times 300 \) (1); 18 000 (J) (1) | 2 | AO2 | Correct answer scores 2. ALLOW 1.8 × 10^4. |
| (b) | correct temperature change 40 (1); \( 18\,000 / (0.50 \times 40) \) (1); 900 with unit J/kg°C (1) | 3 | AO2 | Correct answer with unit scores 3. ALLOW J/kg K. ECF from (a) for the evaluation mark. Unit must be correct for the final mark. |
| (c) | insulate the block / use a lid or lagging (AW) | 1 | AO3 | ALLOW put a little oil in the thermometer hole for better thermal contact. DO NOT ALLOW 'repeat the experiment' or 'use a better thermometer' alone. |
| Total | | 6 | | |

Option 1: \(I = P/V = 12\,000\,000 / 20\,000 = 600\) A. Power wasted \(= I^2R = 600^2 \times 5.0 = 1\,800\,000\) W = 1.8 MW, which is 15% of 12 MW.

Option 2: \(I = 12\,000\,000 / 200\,000 = 60\) A. Power wasted \(= 60^2 \times 5.0 = 18\,000\) W = 18 kW, which is 0.15% of 12 MW.

The current is 10 times smaller at 10 times the potential difference, so the power wasted is 100 times smaller. Option 2 should be chosen as far less energy is wasted as heat in the cables. A step-up transformer at the wind farm raises the potential difference for transmission. A step-down transformer near the users lowers it to a safe value for homes and businesses.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| * | Level 3 (5-6 marks): Calculates the current and the power wasted for both options correctly (or equivalent), compares them, and justifies choosing Option 2 using the link between higher potential difference, lower current and less heating of the cables. Also states a step-up transformer at the wind farm and a step-down transformer at the user end, with their roles. The response is well structured and the reasoning is logical.
Level 2 (3-4 marks): Calculates at least one current and one power loss correctly, or gives a correct qualitative explanation with some calculation. Chooses Option 2 with some reasoning. Mentions transformers but the role or type may be incomplete.
Level 1 (1-2 marks): Some relevant statements, e.g. Option 2 is better, or one correct step of a calculation, or a transformer type named. Little or no reasoning.
Level 0 (0 marks): No response or no response worthy of credit. | 6 | AO2 x3, AO3 x3 | Indicative content: Option 1 current 600 A; Option 2 current 60 A. Power wasted 1.8 MW (1.8 x 10^6 W) for Option 1 and 18 kW (1.8 x 10^4 W) for Option 2. 15% vs 0.15% of output (optional). Higher potential difference gives lower current for the same power. Lower current gives less thermal energy wasted in cables, because loss is proportional to I^2. Option 2 chosen. Step-up transformer at the wind farm; step-down transformer(s) near the consumers to reduce the potential difference to a safe value (about 230 V). ALLOW AW and ORA throughout. ALLOW ECF for a wrong current carried into the power loss calculation. IGNORE comments on cost of cables unless linked to the argument. DO NOT ALLOW energy is 'used up' or 'lost' without reference to being transferred as heat/thermal energy to the surroundings.

(a)(i) A voltmeter is always connected in parallel with the component. (ii) Current heats the wire, which changes its resistance, so the switch is opened to keep the temperature constant.
(b) \(R = V / I = 2.4 / 0.30 = 8.0\ \Omega\).
(c) 1.20 m is 1.20 / 0.50 = 2.4 times longer, so \(R = 8.0 \times 2.4 = 19.2\ \Omega\).
(d) \(P = I^2 R = 0.30^2 \times 8.0 = 0.72\) W.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| (a)(i) | voltmeter | 1 | AO1 | DO NOT ALLOW ammeter |
| (a)(ii) | (to stop) the wire heating up / getting hot AW (1) | 1 | AO2 | ALLOW so resistance does not change / so temperature stays constant. IGNORE to save the cell / for safety only |
| (b) | \(R = V/I\) or 2.4 / 0.30 (1); 8.0 (\(\Omega\)) (1) | 2 | AO2 | Correct answer scores 2 marks. ALLOW 8 |
| (c) | 1.20 / 0.50 or 2.4 or 8.0 / 0.50 = 16 (\(\Omega\) m\(^{-1}\)) (1); 19.2 (\(\Omega\)) (1) | 2 | AO2 | Correct answer scores 2 marks. ALLOW 19 or 20 (rounded). ECF from (b) |
| (d) | 0.72 (W) | 1 | AO2 | ALLOW 0.7. ALLOW ECF from (b) using \(0.30^2 \times\) their R. Unit W required only if not printed |
| Total | | 7 | | |

(a) Wrap the beaker in insulation (or use a lid).
(b) \(\Delta E = mc\Delta\theta = 0.50 \times 4200 \times 18 = 37\,800\) J.
(c) Energy supplied = \(P t = 60 \times 700 = 42\,000\) J. Efficiency = useful energy output / total energy input = \(37\,800 / 42\,000 = 0.90\), so 90 %.
(d) Stirring makes the temperature uniform, so the thermometer reading is representative of the whole of the water.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| (a) | insulation / lagging / lid / cover the beaker (AW) | 1 | AO2 | ALLOW cotton wool / foam / lower-conductivity material around beaker. IGNORE 'use a bigger heater'. |
| (b) | substitution \(0.50 \times 4200 \times 18\) | 1 | AO2 | Equation given, so mark substitution. |
| (b) | evaluation = 37 800 | 1 | AO2 | ALLOW 38 000 (2 s.f.) or 37 800 without working for 2 marks. |
| (b) | unit J | 1 | AO2 | ALLOW kJ if consistent (37.8 kJ). |
| (c) | energy supplied = \(60 \times 700\) = 42 000 J | 1 | AO2 | Energy = power × time. |
| (c) | efficiency = useful output / total input = 37 800 / 42 000 (ECF from (b)) | 1 | AO2 | |
| (c) | 90 (%) | 1 | AO2 | ALLOW 0.9 as 1 mark only if no percentage given. ECF from (b). |
| (d) | so the temperature is the same throughout the water / the thermometer reads the average temperature / thermal energy is spread evenly (AW) | 1 | AO3 | IGNORE 'to make it a fair test' without explanation. |

(a) A battery gives a steady current, so the magnetic field in the iron core does not change. Electromagnetic induction needs a changing magnetic field, so no p.d. is induced in the secondary coil.

(b) \( V_s = V_p \times N_s / N_p = 230 \times 60 / 1200 = 11.5 \) V.

(c) \( V_pI_p = V_sI_s \), so \( I_p = (11.5 \times 2.0)/230 = 23/230 = 0.10 \) A.

(d) Input power = \( 230 \times 0.12 = 27.6 \) W. Output power = \( 11.5 \times 2.0 = 23 \) W. Efficiency = \( 23/27.6 \times 100 = 83.3\% \), which is 83% to 2 significant figures.

(e) Some energy is dissipated as thermal energy because of the resistance of the coils or eddy currents in the core, or some magnetic flux does not link the secondary coil.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| (a) | d.c. gives a constant/steady current (in the primary) (1); so the magnetic field/flux in the core does not change, so no p.d./current is induced in the secondary (1) | 2 | AO1.1 x1, AO2.1 x1 | ALLOW no changing magnetic field. ALLOW no electromagnetic induction (only with the idea of no change). DO NOT ALLOW 'd.c. does not flow through a transformer' without reference to the field. |
| (b) | \( V_s = 230 \times 60 / 1200 \) (1); 11.5 (V) (1) | 2 | AO2.1 x2 | Correct answer scores 2. ALLOW 12 (V) for 1 mark only if working is shown. 1 mark for correct substitution or rearrangement. |
| (c) | \( V_pI_p = V_sI_s \) used, or \( 230 \times I_p = 11.5 \times 2.0 \) (1); \( I_p = (11.5 \times 2.0)/230 \) (1); 0.10 (A) (1) | 3 | AO2.1 x3 | Correct answer scores 3. ECF from the answer to (b). ALLOW 0.1. |
| (d) | efficiency = useful output / input power (\( \times 100 \)) with 23 W and 27.6 W, or \( 23 / 27.6 \) (1); 83 (%) (1) | 2 | AO2.1 x2 | Correct answer of 83 scores 2. ALLOW 83.3 for 1 mark if not rounded to 2 s.f. ALLOW 0.83 for 1 mark only (no %). ECF from (b). |
| (e) | any one from: heating of the coils (due to resistance) / eddy currents heating the iron core / magnetic flux not all linking with the secondary coil / vibration or sound (1) | 1 | AO3.3b | AW. IGNORE 'friction' and 'energy is lost'. DO NOT ALLOW 'energy is destroyed'. |
| Total | | 10 | | |

(a) Vs = Vp x Ns / Np = 230 x 60 / 1200 = 11.5 V.
(b) A battery supplies direct current, so the current in the primary coil is constant and produces a constant magnetic field. The magnetic flux linking the secondary coil does not change, so no potential difference is induced in it.
(c) Vp Ip = Vs Is, so Ip = (11.5 x 2.0) / 230 = 0.10 A.
(d) F = B I l = 0.040 x 3.0 x 0.12 = 0.0144 N (1.4 x 10^-2 N).
(e) Reverse the direction of the current; reverse the direction of the magnetic field (swap the magnet poles).

| Q | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| (a) | Substitution / rearrangement: Vs = 230 x 60 / 1200 (1); 11.5 (V) (1) | 2 | AO2 | ALLOW 12 V (rounded) for the second mark. Correct answer only gets 2 marks. |
| (b) | A battery supplies d.c. / constant current (1); so the magnetic field / flux in the core does not change, so no p.d. is induced in the secondary coil (1) | 2 | AO1 x1, AO2 x1 | ALLOW 'no changing field' AW. IGNORE 'battery is too weak'. DO NOT ALLOW 'current cannot flow through the core'. |
| (c) | Vp Ip = Vs Is used / 230 x Ip = 11.5 x 2.0 (1); rearrangement or Ip = 23 / 230 (1); 0.10 (A) (1) | 3 | AO2 | ECF from (a). ALLOW 0.1. ALLOW 0.104 if 12 V used from (a). |
| (d) | F = B x I x l selected / substitution 0.040 x 3.0 x 0.12 (1); 0.0144 (1); unit N (1) | 3 | AO2 | ALLOW 0.014 or 1.4 x 10^-2. Unit mark can be gained only with a numerical answer in newtons. |
| (e) | Reverse the current (1); reverse the magnetic field / swap the poles of the magnet (1) | 2 | AO1 | ALLOW reverse the battery / cell connections. DO NOT ALLOW 'increase the current', 'turn the wire round' or 'use a stronger magnet'. |
| Total | | 12 | | |

(a) An a.c. supply makes the current in the primary coil change continuously, so the magnetic field in the iron core keeps changing. The changing field through the secondary coil induces a potential difference (electromagnetic induction). With d.c. the field is constant after switching on, so no potential difference is induced.
(b) \( V_s = V_p \times N_s / N_p = 16 \times 150 / 1200 = 2.0 \) V.
(c) The secondary p.d. is smaller than the primary p.d., so it is a step-down transformer.
(d)(i) \( P = VI = 2.0 \times 0.60 = 1.2 \) W.
(d)(ii) \( I_p = V_s I_s / V_p = 1.2 / 16 = 0.075 \) A.
(e) Keep the primary coil, the a.c. input potential difference and the core unchanged. Change the number of secondary turns in steps (for example 50, 100, 150, 200, 250). Measure the secondary p.d. each time with an a.c. voltmeter. Repeat and average, then compare.
(f) For a given power, a higher potential difference means a smaller current (P = VI). Heating loss in the cables is proportional to I squared R, so less energy is wasted as thermal energy, increasing efficiency.

| Q | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| (a) | changing current / a.c. in primary produces changing magnetic field (in core) (1); changing field induces a (changing) p.d. in secondary coil / d.c. gives constant field so no p.d. induced (1) | 2 | 1.1 x2 | ALLOW flux for field. IGNORE just 'a.c. changes direction' |
| (b) | Vs = Vp x Ns / Np or correct substitution (1); 2.0 (V) (1) | 2 | 2.2 | Correct answer gets both marks. ALLOW 2 |
| (c) | step-down | 1 | 1.2 | Only one box ticked |
| (d)(i) | P = V x I or 2.0 x 0.60 (1); 1.2 (W) (1) | 2 | 2.2 | ALLOW 1.2 W with unit |
| (d)(ii) | Ip = Vs x Is / Vp or 1.2 / 16 (1); 0.075 (A) (1) | 2 | 2.2 | ECF from (d)(i). ALLOW 7.5 x 10^-2 |
| (e) | change number of secondary turns / use different secondary coils (1); measure secondary p.d. with (a.c.) voltmeter each time (1); control the primary turns / input p.d. / same core (1) | 3 | 3.3a, 3.3b | ALLOW repeat readings and average for a mark in place of a control. DO NOT ALLOW d.c. voltmeter |
| (f) | higher p.d. gives lower current (for same power) (1); less energy wasted / thermal energy lost in cables (1); because heating is proportional to I squared (x R) / increases efficiency of transmission (1) | 3 | 1.1, 2.1 | AW. IGNORE 'cheaper' unless linked to energy loss |
| Total | | 15 | | |

An irregular solid has no simple dimensions, so its volume is found by displacement. The rise in water level in the measuring cylinder equals the volume of the stone.

B1 for B. Do not allow A (shape is irregular), C (gives mass, not volume) or D (time does not give volume).

\(V_p/V_s = N_p/N_s\), so \(N_s = N_p \times V_s/V_p = 1150 \times 23/230 = 115\).

B1 for B (115). Method: correct use of turns ratio. Do not allow 11 500 (ratio inverted).

\(F = BIl = 0.20 \times 3.0 \times 0.15 = 0.090\,\text{N}\).

B1 for C. Method: F = B x I x l with correct substitution; accuracy: 0.090 N. Powers-of-ten errors give A or D.

For a given power, \(P = VI\), a higher potential difference means a lower current. Heating loss depends on \(I^2R\), so less energy is dissipated in the cables.

B1 for D. Do not allow A (current is reduced, not increased) or B (the Grid uses a.c.).

Efficiency = useful output / total input = 600 / 800 = 0.75 = 75%.

B1 for C. Do not allow 133% (ratio inverted) or 25% (wasted fraction).

Volume = 0.20 x 0.10 x 0.050 = 0.0010 m^3. Density = mass / volume = 2.7 / 0.0010 = 2700 kg/m^3.

B1 for C. Method: volume calculated correctly, then mass divided by volume. Powers-of-ten slips give A, B or D.

Total resistance in series = 4.0 + 8.0 = 12 ohms. \(I = V/R = 12/12 = 1.0\,\text{A}\).

B1 for B. Method: add resistances, then I = V / R. Do not allow 3.0 A (only the 4.0 ohm resistor used) or 1.5 A (only the 8.0 ohm resistor used).

Light from distant galaxies is red-shifted: its wavelength is increased because the galaxies are moving away from us. This shows space is expanding.

B1 for A. Do not allow B (blue-shift would show approach).

Volume of stone = 110 − 50 = 60 cm\(^3\). Density = mass / volume = \(156 \div 60 = 2.6\) g/cm\(^3\).

B (1 mark). A uses 156/110; C uses 156/50 (volume of water only); D is volume/mass inverted. AO2.

Timing many swings makes the reaction-time error a small fraction of the total time. Dividing by the number of swings gives the period.

B (1 mark). A and D leave the reaction-time error large compared with the time measured; C is the wrong calculation. AO3.

\(V_p/V_s = N_p/N_s\), so \(N_s = 2400 \times 15 \div 120 = 300\).

B (1 mark). A inverts the ratio; C and D use subtraction. AO2.

\(F = BIl = 0.40 \times 3.0 \times 0.25 = 0.30\) N.

B (1 mark). C omits the length; D divides by B and l instead of multiplying; A divides by the current. AO2.

With the switch in the live wire, opening it disconnects the appliance from the live (about 230 V) supply, so touching internal parts is not dangerous.

A (1 mark). B is wrong because the neutral wire completes the circuit; C is wrong because the earth wire carries current only in a fault; D is wrong because a switch does not change a.c. to d.c. AO1.

Efficiency = useful output / input = \(2 \div 8 = 0.25\), which is 25%.

B (1 mark). C is the wasted fraction; D inverts the ratio; A is not derived from the data. AO2.

Time = 2.5 × 60 = 150 s. \(Q = It = 0.40 \times 150 = 60\) C.

B (1 mark). A forgets to convert minutes to seconds; D is 150/0.40; C divides 0.40 by 150. AO2.

(a) \( E = Pt = 50 \times 300 = 15\,000 \) J.
(b) Measure the mass of the block with a balance and the initial temperature. Switch on the heater and a stopwatch, and heat for a fixed time, then record the highest temperature reached (it may continue to rise after switching off). Calculate the temperature change. Energy supplied = Pt (or joulemeter reading). Rearrange \( \Delta E = mc\Delta T \) to \( c = \Delta E / (m\Delta T) = 15000 / (1.0 \times 17) \approx 880 \) J/(kg °C). Some energy is transferred to the surroundings, so the energy actually stored in the block is less than 15 000 J and the calculated c is too high. Improve by adding more insulation, a few drops of oil in the thermometer hole for good thermal contact, heating for a shorter time over a smaller rise, repeating the experiment and calculating a mean, or plotting a graph of energy against temperature and using the gradient.

Q2(a) [AO2] 2 marks: substitution \( 50 \times 300 \) (1); 15 000 (J) (1). ALLOW 1.5 × 10^4. Correct answer only = 2 marks.
Q2(b)* [AO1 x2, AO2 x2, AO3 x2] Level of response, best-fit.
Level 3 (5–6 marks): Complete, logical plan including measurement of mass and temperatures with named apparatus, correct use of energy = Pt and \( \Delta E = mc\Delta T \) (or rearranged) with the value ≈ 880 J/(kg °C), AND a clear explanation of the effect of energy loss (measured c is too high / energy in block less than supplied) with at least one valid improvement. Ideas are linked in a clear sequence.
Level 2 (3–4 marks): Describes a method with some measurements and uses at least one correct equation, with partial reference to energy loss or improvements; may lack the calculation or a link between energy loss and the result. Some logical structure.
Level 1 (1–2 marks): Simple statements, e.g. measures temperature before and after heating or mentions insulation, with little or no link to the calculation. Limited structure.
Level 0 (0 marks): No response or no creditworthy content.
Indicative content: balance for mass; thermometer for start/final temperature; stopwatch for time; maximum temperature reached; energy = power × time; \( c = E/(m\Delta T) \); c ≈ 880 J/(kg °C); heat loss to surroundings means c calculated is too high (ORA); more/better insulation; oil in thermometer hole; repeat and calculate a mean; graph of E against \( \Delta T \) and use gradient / m. ALLOW AW. IGNORE references to human error or reading the thermometer at eye level only. DO NOT ALLOW c calculated is too low due to energy loss.

(a) Time = 5.0 × 60 = 300 s. Energy = 40 × 300 = 12 000 J.

(b) Temperature change = 45 − 20 = 25 °C. Rearranging, c = \( \frac{\Delta E}{m \Delta \theta} \) = 12 000 / (0.50 × 25) = 960 J/(kg °C). This is about 7% higher than 900 J/(kg °C). In reality some energy is transferred to the surroundings (and to the heater and thermometer), so less than 12 000 J raised the block's temperature. The measured temperature rise is therefore too small, and the calculated c is too large. Improvements: add more or thicker insulation; measure the highest temperature reached after switching off, since the thermometer reading lags; add a little oil in the thermometer hole to improve thermal contact; stir/wait for thermal equilibrium; repeat and average; use a joulemeter or ammeter and voltmeter for the actual energy supplied.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| (a) | Correct substitution of 40 × 300 (time converted to seconds) | 1 | AO2 | ALLOW 40 × 5 × 60. DO NOT ALLOW 40 × 5.0 (= 200). |
| (a) | 12 000 (J) | 1 | AO2 | ALLOW 1.2 × 10^4. ALLOW 1 mark for 200 J if the time is not converted (ECF is not available for the final mark). Correct answer scores 2 marks. |
| (b)* | Level 3 (5–6 marks): correct calculation of c = 960 (J/(kg °C)) with working; comparison with 900 stated as higher; clear explanation linked to energy transferred to surroundings so the temperature rise is too small; two sensible procedural improvements described. The response is well structured and logically linked. | 5–6 | AO2 AO3 | Calculation correct and both improvements and explanation present gives 6; one weak link gives 5. |
| (b)* | Level 2 (3–4 marks): c calculated (correctly, or with one error giving ECF), and either a reasoned explanation of the difference or improvements described, with some links between ideas. | 3–4 | AO2 AO3 | e.g. correct 960 with comparison and one improvement = 3. |
| (b)* | Level 1 (1–2 marks): a partial attempt, e.g. a correct substitution or one relevant statement about energy loss or insulation, with little or no linking. | 1–2 | AO1 AO2 | |
| (b)* | Level 0 (0 marks): no response or no response worthy of credit. | 0 | | |
| Indicative content for (b) | Rearrangement c = ΔE / (m × Δθ); Δθ = 25 °C; c = 12 000 / (0.50 × 25) = 960 J/(kg °C) (ALLOW 960 to 2 or more significant figures); 960 is higher than 900 (about 7%); some energy is transferred (dissipated) to the surroundings / wasted / heats the heater and thermometer; so the temperature rise is smaller than if all energy stayed in the block; so the calculated c is too large. Improvements: more/thicker insulation (AW); use lid or insulate the top; add oil/water in the thermometer hole for good thermal contact; record the maximum temperature after switching off; repeat the experiment and calculate a mean; use a joulemeter or voltmeter and ammeter for the energy supplied. | | | IGNORE human error, 'faulty thermometer'. DO NOT ALLOW 'energy is lost' or 'used up' without reference to being transferred/dissipated to the surroundings (AW). |

(a) F = BIl = 0.080 × 4.5 × 0.12 = 0.0432 N ≈ 0.043 N.
(b) Increase the current (larger p.d. supply); use stronger magnets to increase B.
(c)(i) \(V_s = V_p \times N_s / N_p = 230 \times 46 / 920 = 11.5\) V.
(c)(ii) \(V_p I_p = V_s I_s\) so \(I_p = 11.5 \times 3.0 / 230 = 0.15\) A.

(a) Substitution 0.080 × 4.5 × 0.12 (1); 0.0432 or 0.043 (1); N (1). ALLOW ECF for unit only if the numerical value is correct. DO NOT ALLOW newtons written as 'n'.
(b) Any two from: increase the current / p.d. (1); use a stronger magnet / increase magnetic flux density (1); increase the length of wire in the field is NOT creditable as wire is unchanged. IGNORE 'increase the voltage' alone if no link to current (AW).
(c)(i) Correct rearrangement/substitution 230 × 46 ÷ 920 (1); 11.5 (V) (1). ALLOW 12 V.
(c)(ii) Correct equation substitution 230 × Ip = 11.5 × 3.0 (1); 0.15 (A) (1). ALLOW ECF from (c)(i). ALLOW 0.150.

(a) E = Pt = 40 × 300 = 12 000 J. ΔT = 36 − 19 = 17 °C. c = E / (mΔT) = 12 000 / (0.80 × 17) = 882 ≈ 880 J/kg°C.
(b) Method: measure the mass of the block with a balance; record the initial temperature; switch on the heater and start a stopwatch; measure the power (or the current and p.d.); switch off after a fixed time and record the highest temperature reached, since thermal energy continues to spread through the block; repeat and average. Improvement: some energy is transferred to the surroundings so less energy than 12 000 J warms the block; the calculated c is therefore greater than the true value. Wrap the block in insulation (e.g. a lagging layer or foam) and cover the top; put a little oil in the thermometer hole to improve thermal contact; measure the energy using a joulemeter.

(a) E = 40 × 300 = 12 000 J (1); correct substitution into c = E/(mΔT), ΔT = 17 (1); 880 (J/kg°C) (1). ALLOW 882 or 880. ALLOW ECF for 300 s error in E.
(b) Levels of response, 6 marks.
Level 3 (5–6 marks): A coherent, detailed method that includes at least three valid steps (measuring mass, recording start and highest temperature, timing/power, repetition) AND a linked explanation of energy loss to the surroundings and at least one improvement (insulation, lid, oil in the hole, joulemeter) with its effect on the result (c overestimated because the temperature rise is smaller than expected/AW). Logical sequence with correct scientific terms.
Level 2 (3–4 marks): A method with at least two correct steps, and a basic statement of an improvement or energy loss, with partial linking to the result. Some structure.
Level 1 (1–2 marks): One or two isolated points about procedure or energy loss; limited detail or logic.
Level 0 (0 marks): No relevant content.
Indicative content: balance for mass; thermometer in the block; stopwatch; record the maximum temperature after switching off; repeat and average; some energy transferred to surroundings; measured c is higher than the accepted value; insulate; reduce the time of heating; oil for thermal contact; measure p.d. and current to obtain the actual power. IGNORE 'the heater is faulty'. DO NOT ALLOW insulation to 'stop all' energy loss.

(a) The power transmitted is the product of p.d. and current. Raising the p.d. lowers the current for the same power. The power wasted in the cable is I²R, so a smaller current greatly reduces energy dissipated as thermal energy, making transmission more efficient.
(b) I = P / V = 50 000 000 / 400 000 = 125 A.
(c) P = I²R = 125² × 8.0 = 15 625 × 8.0 = 125 000 W.
(d) Useful output = 50 000 000 − 125 000 = 49 875 000 W. Efficiency = 49 875 000 / 50 000 000 × 100 = 99.75 % ≈ 99.8 %. (Using 99.75 gives 99.8 to 3 s.f.)

(a) Higher potential difference means lower current (for the same power) (1); the power loss/heating in the cables depends on the current squared (I²R) / lower current means less thermal energy dissipated (1); so the transmission is more efficient / more power reaches consumers (1). IGNORE 'to make it safer'. DO NOT ALLOW 'resistance decreases'.
(b) 50 000 000 ÷ 400 000 (1); 125 (A) (1).
(c) 125² × 8.0 (1); 125 000 (W) / 1.25 × 10^5 (1). ALLOW ECF from (b). ALLOW 125 kW.
(d) (50 000 000 − 125 000) ÷ 50 000 000 × 100 (1); 99.8 (%) (1). ALLOW 99.75. ALLOW ECF from (c). DO NOT ALLOW 100 %.

(a) Gas pressure is caused by particles colliding with the container walls. With a smaller volume, the particles are closer together and collide with the walls more frequently, so the total force per unit area increases.
(b) \(p_1V_1 = p_2V_2\), so \(p_2 = 100 \times 60 / 24 = 250\) kPa.
(c) The plunger does work on the gas. This energy is transferred to the kinetic energy of the particles faster than it can be transferred to the surroundings, so the internal energy and temperature increase.
(d) pV is constant so p is inversely proportional to V: a curve falling as V increases, never touching the axes.

(a) Particles collide with the walls of the container (1); in a smaller volume the collisions are more frequent / particles hit the walls more often (1); greater (total) force per unit area / so pressure increases (1). DO NOT ALLOW 'particles move faster' or 'particles collide with each other more'.
(b) Correct equation used: 100 × 60 = p × 24 (1); rearrangement 100 × 60 ÷ 24 (1); 250 (kPa) (1). ALLOW 250 000 Pa. ALLOW 2 marks for 40 (inverted ratio).
(c) Work is done on the gas (1); energy transferred to the kinetic energy (store) of the particles / internal energy increases, so temperature rises (1). IGNORE 'friction'.
(d) Second box ticked (1).

(a) With d.c. the current in the primary is steady, so the magnetic field in the iron core is constant. The magnetic flux linking the secondary coil does not change, so no potential difference is induced (electromagnetic induction needs a changing field).

(b) \( N_s = N_p \times V_s / V_p = 1150 \times 12 / 230 = 60 \) turns.

(c) Power in secondary: \( V_s I_s = 12 \times 2.5 = 30 \) W. For 100% efficiency \( V_p I_p = 30 \), so \( I_p = 30 / 230 = 0.1304 \) A = 0.13 A (2 s.f.).

(d) Connect the primary coil to a low-voltage a.c. supply and wind a secondary coil on the same core. Use an a.c. voltmeter across the secondary and record Vs. Change the number of secondary turns (e.g. 10, 20, 30, 40, 50), measuring Vs each time. Keep the primary p.d. (and number of primary turns, and the core) the same.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| (a) | d.c. produces a constant / steady magnetic field (in the core) (1); no changing magnetic flux (linking the secondary) so no p.d. induced (1) | 2 | AO1.1 / AO2.1 | ALLOW AW 'field does not change / is not alternating'. DO NOT ALLOW 'no current' in the primary. IGNORE references to heating. |
| (b) | correct selection / rearrangement: Ns = Np x Vs / Vp (1); substitution 1150 x 12 / 230 (1); 60 (turns) (1) | 3 | AO2.1 x2, AO2.2 | Correct answer with no working scores 3. ALLOW 1 mark for 5 / 0.0167 if equation inverted (ECF substitution only). |
| (c) | Vs x Is = 12 x 2.5 = 30 (W) (1); Ip = 30 / 230 (1); 0.13 (A) (1) | 3 | AO2.1, AO2.2 | ALLOW 0.130. Correct answer with no working scores 3. If not 2 s.f. but 0.1304... shown, max 2. ALLOW ECF from an incorrect Vs x Is. |
| (d) | connect primary to a.c. supply and measure Vs using an (a.c.) voltmeter across the secondary (1); change the number of secondary turns / repeat with different Ns and record Vs each time (1); keep one variable constant, e.g. primary p.d. / Np / frequency / same core (1) | 3 | AO3.3 | ALLOW AW. DO NOT ALLOW 'keep the voltage the same' without specifying primary. DO NOT ALLOW ammeter for measuring Vs. |

(a) The p.d. increases from 11 000 V to 132 000 V, so the transformer is step-up.

(b) \(\frac{V_p}{V_s}=\frac{N_p}{N_s}\), so \(N_s = N_p \times \frac{V_s}{V_p} = 550 \times \frac{132\,000}{11\,000} = 550 \times 12 = 6600\) turns.

(c) For a 100% efficient transformer, power in = power out = 33 MW = 33 000 000 W.
Primary: \(I_p = \frac{P}{V_p} = \frac{33\,000\,000}{11\,000} = 3000\) A.
Secondary: \(I_s = \frac{33\,000\,000}{132\,000} = 250\) A.

(d) Power = potential difference × current, so for the same power a higher p.d. means a lower current. The power wasted in the cables is \(I^2R\), so a smaller current gives much less heating (energy wasted) in the cables of fixed resistance.

(e) \(P = I^2R = 250^2 \times 4.0 = 62\,500 \times 4.0 = 250\,000\) W = 250 kW.

| Q | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| (a) | step-up box ticked | 1 | AO1 | Only one box ticked. |
| (b) | \(N_s/N_p = V_s/V_p\) or \(550 \times 132000/11000\) (1); ratio 12 or \(132000/11000\) evaluated (1); 6600 (1) | 3 | AO2 | Full marks for correct answer with no working. ALLOW 1 mark for rearrangement alone. |
| (c) | \(I = P/V\) or \(V_pI_p = V_sI_s\) (1); primary 3000 A (1); secondary 250 A (1) | 3 | AO2 | ALLOW ECF for a power conversion error (e.g. 33 MW taken as 33000 W) once only. Correct answers with no working get full marks. |
| (d) | P = VI / high p.d. gives a lower current for the same power (1); wasted power depends on \(I^2\) / heating depends on current squared (1); so less thermal energy / heating in cables (AW) (1) | 3 | AO1, AO2 | IGNORE references to thinner cables. DO NOT ALLOW 'resistance of cables decreases'. |
| (e) | \(P = I^2R\) with substitution \(250^2 \times 4.0\) (1); 250 (kW) (1) | 2 | AO2 | ALLOW 250 000 W for 1 mark only if kW conversion not shown. ALLOW ECF from (c) secondary current. |