2025 OCR GCSE Gateway Science - Physics A - J249 Practice Paper with Answers

For a given power, \(P = VI\), so a higher potential difference means a smaller current. A smaller current heats the cables less, so less energy is wasted.

B = 1 mark. A: wrong, current decreases. C: transformers do not change a.c. to d.c. D: high voltage cables are not safe to touch.

\(I = V \div R = 12 \div 4 = 3\) A.

C = 1 mark. A comes from \(R \div V\). B comes from \(V \times R\). D comes from \(V - R\).

\(\text{density} = \text{mass} \div \text{volume} = 0.45 \div 0.0005 = 900\) kg/m\(^3\).

A = 1 mark. B comes from \(\text{volume} \div \text{mass}\). C and D are arithmetic or unsupported values.

\(\text{efficiency} = \text{useful output} \div \text{total input} = 150 \div 200 = 0.75 = 75\%\).

D = 1 mark. A is the wasted fraction (50 J out of 200 J). C comes from inverting the ratio. Efficiency cannot be above 100%.

Timing many swings makes the reaction-time error a much smaller fraction of the total time. Dividing by 20 then gives the time for one swing.

C = 1 mark. A: too few readings, and each one has a large percentage error. B: the stopwatch resolution is not the main source of error. D: multiplying gives a wrong value.

With the switch in the neutral wire, the live wire (at about 230 V) stays connected to the kettle even when the switch is off. Touching a faulty part could cause a current through a person to earth.

B = 1 mark. A: the kettle would still work when the switch is closed. C and D are not consequences of the switch position.

\(\lambda = v \div f = 340 \div 170 = 2\) m.

D = 1 mark. A comes from \(f \div v\). C comes from \(v \times f\). B is not supported by the calculation.

\(k = F \div x = 6 \div 0.03 = 200\) N/m.

A = 1 mark. B comes from \(x \div F\). C and D come from multiplying \(F \times x\) (0.18) or a related slip.

Power = potential difference × current. A step-up transformer raises the potential difference, so for the same power the current is smaller. Less current means less energy is transferred as heat in the cables. The answer is B.

B [1 mark]. A: cable thickness is not changed by the transformer. C: transformers do not increase power (power is conserved apart from small losses). D: frequency stays at 50 Hz and does not affect wave or energy speed.

\(I = \frac{V}{R} = \frac{12}{40} = 0.30\) A. The answer is C.

C [1 mark]. A comes from dividing resistance by potential difference (R ÷ V). B comes from multiplying V × R. D comes from adding the values.

\(\text{density} = \frac{\text{mass}}{\text{volume}} = \frac{540}{200} = 2.7\) g/cm\(^3\). The answer is A.

A [1 mark]. B is volume ÷ mass (inverted). C is mass − volume. D is mass × volume.

One sheet is thinner than the smallest ruler division, so it cannot be measured directly. Measuring a large stack gives a measurable thickness. Dividing by the number of sheets gives the thickness of one sheet. The answer is D.

D [1 mark]. A and B cannot resolve a thickness smaller than the ruler scale. C uses the wrong operation.

\(\text{efficiency} = \frac{\text{useful output energy}}{\text{input energy}} = \frac{600}{800} = 0.75 = 75\%\). The answer is B.

B [1 mark]. A is the proportion wasted (200 ÷ 800). C is input ÷ output. D is a conversion error giving 0.75 as a percentage.

Current from the battery splits between the two parallel branches. The lamps are identical so the current divides equally: 0.60 ÷ 2 = 0.30 A. The answer is A.

A [1 mark]. B assumes the whole current passes through each lamp. C doubles the current. D halves the current twice.

The live wire carries current to the appliance. The neutral wire completes the circuit by returning current to the supply. The earth wire carries current only in a fault. The answer is D.

D [1 mark]. A carries current to the appliance at about 230 V. B is a safety wire that carries no current in normal use. C is a safety device, not a wire in the cable.

(a) E = P × t = 50 × 200 = 10 000 J.
(b) 40 − 18 = 22 °C.
(c) Oil fills the air gap, so thermal energy transfers between the block and thermometer better and the reading is accurate.
(d) Less energy is lost to the surroundings, so more of the heater's energy raises the block's temperature.
(e) \( \Delta Q = m c \Delta\theta \) so \( c = \Delta Q / (m \Delta\theta) = 10\,000 / (0.50 \times 22) = 909.09 \approx 910 \) J/kg °C.
(f) Some energy is still lost to the surroundings, so the energy actually gained by the block is less than 10 000 J; using 10 000 J makes the calculated value too large.

(a) [2] Correct answer 10 000 (J) = 2 marks. If wrong: substitution 50 × 200 = 1 mark. ALLOW 1.0 × 10^4. ALLOW 10 kJ.
(b) [1] 22 (°C). ALLOW 22 K.
(c) [1] Any one from: improves thermal contact / better conduction between block and thermometer; removes air gap (air is an insulator); so thermometer reads the block's temperature accurately. IGNORE 'to make it a fair test'.
(d) [1] Any one from: reduces energy loss to the surroundings; more of the energy from the heater stays in the block / gives a more accurate temperature rise. ALLOW 'less heat escapes'.
(e) [4] Final answer 910 (J/kg °C) with unit = 4 marks (909 with unit = 3 marks as not rounded to 2 s.f.).
Otherwise: M1 rearranges to c = ΔQ/(mΔθ) (can be implied by substitution); M2 substitutes 10 000 / (0.50 × 22) (ECF from (a) and (b)); M3 evaluates correctly (909.09...) and rounds to 2 s.f. (ECF); M4 unit J/kg °C (ALLOW J/kg K, J kg⁻¹ °C⁻¹). Unit mark awarded independently of the numerical answer. DO NOT ALLOW J/kg alone.
(f) [1] Any one from: some energy still transferred to the surroundings / insulation not perfect; block did not reach the full temperature in the thermometer region (uneven heating), so the temperature rise is lower than expected; energy used to heat the heater/thermometer/insulation. IGNORE 'human error' and 'faulty equipment'.

(a) Energy transferred: \( E = Pt = 40 \times 300 = 12\,000 \text{ J} \).
Temperature change = 42 − 18 = 24 °C.
Using \( \Delta Q = mc\Delta\theta \), \( c = \dfrac{\Delta Q}{m\Delta\theta} = \dfrac{12\,000}{0.50 \times 24} = 1000 \text{ J/kg °C} \).
(b) Some energy is lost to the surroundings, so the block gains less energy than the heater supplies. The calculation uses too large an energy value, giving a higher c.
(c) Wrap the block in thermal insulation (e.g. a layer of foam or cloth).
(d) The oil fills air gaps so there is good thermal contact between the block and thermometer, giving a more accurate temperature reading.
(e) Allow the block and heater to cool before touching them.
(f) Repeat the investigation and calculate a mean value.

(a) Correct final answer of 1000 (J/kg °C) scores 3 marks.
Otherwise:
- energy = 40 × 300 = 12 000 (J) [1]
- substitution into c = E / (m × Δθ), i.e. 12 000 / (0.50 × 24) [1] (temperature change 24 must be used, not 42 or 18)
- 1000 [1]
ALLOW 1 mark for 12 000 / (0.50 × 42) or similar if temperature change is incorrect (ECF applies to final mark only).
ALLOW 1.0 × 10^3. DO NOT ALLOW the answer without working if it is incorrect.

(b) Any one from:
- some energy was transferred/lost to the surroundings [1]
- the block received less energy than the heater supplied [1]
IGNORE 'heat loss' without reference to the surroundings or to the block receiving less energy. DO NOT ALLOW 'the heater is not 40 W' without justification.

(c) Any one from:
- wrap/surround the block in insulation (e.g. foam, cotton wool, lagged jacket) [1]
- place a lid/insulating layer on top of the block [1]
DO NOT ALLOW 'put it in a box' or 'use a different material' without insulation.

(d) Any one from:
- improves thermal contact between block and thermometer [1]
- removes/fills air gaps (air is an insulator) so the thermometer reading is closer to the block temperature [1]
IGNORE 'to make it a fair test'.

(e) Any one from:
- wait for the block/heater to cool before touching/moving it [1]
- use tongs/heat-resistant gloves to handle the block [1]
IGNORE 'wear goggles' on its own.

(f) Any one from:
- repeat the experiment and calculate a mean [1]
- repeat the experiment and discard anomalous results [1]
- take more temperature readings over time and plot a graph to find the gradient [1]
IGNORE 'repeat it' without reference to a mean or comparison. IGNORE 'use a more accurate thermometer'.

Total: 8 marks (AO1 2, AO2 3, AO3 3).

(a) The wind farm output goes to a step-up transformer; a step-down transformer is used near homes.
(b) Wind is naturally replenished and will not run out.
(c) \( I = P \div V = 6\,900\,000 \div 23\,000 = 300 \) A.
(d) \( P = I^2 R = 300^2 \times 2.0 = 90\,000 \times 2.0 = 180\,000 \) W.
(e) For the same power, \( P = VI \) means a higher potential difference gives a smaller current. Power wasted is \( I^2 R \), so a smaller current gives much less heating of the cables.

(a) step-up (1); step-down (1). Both must be in the correct order. DO NOT ALLOW if both words are the same.
(b) Any one from: it can be replaced as fast as it is used / it will not run out / it is not used up / it comes from a source that is continually replenished (1). IGNORE 'it is clean' or 'it is free'. DO NOT ALLOW 'it can be reused'.
(c) Final answer 300 (A) = 2 marks. If the answer is wrong: correct rearrangement \( I = P/V \) or correct substitution 6 900 000 / 23 000 = 1 mark. ALLOW 0.3 kA.
(d) Final answer 180 000 (W) = 2 marks. ALLOW 180 kW or 1.8 × 10^5 W. If the answer is wrong: correct substitution \( 300^2 \times 2.0 \) = 1 mark. ALLOW 1 mark for 90 000 (current squared only).
(e) Higher potential difference gives a lower current (for the same power) (1). Less thermal energy / heating (power loss) in the cables because power wasted depends on current squared / \( I^2R \) (1). IGNORE 'the resistance is lower'. DO NOT ALLOW 'less energy is made' or 'voltage is reduced'.

Power transmitted is \(P = VI\). A step-up transformer at the wind farm increases the potential difference to a very high value. For the same power, the current in the cables is therefore much smaller. The power wasted in the cables is \(P = I^2R\), so a smaller current gives much less wasted power (halving the current cuts the wasted power to a quarter). Less energy is transferred as thermal energy to the surroundings, so more of the generated energy reaches the town and the system is more efficient. Very high potential differences are dangerous and unsuitable for homes and appliances, so a step-down transformer near the town reduces the potential difference to a safe level, about 230 V, for domestic use. Transformers only work with alternating current, which is what the National Grid uses.

Level of response (best fit). Scientific coverage decides the level; clarity, logical order and use of evidence decide the mark within the level.

Level 3 (5-6 marks): Gives a clear, connected explanation covering all three strands: (1) step-up transformer raises potential difference and lowers current for the same power; (2) lower current means less energy wasted as heat in the cables, using I^2R or equivalent; (3) step-down transformer in the town reduces the potential difference to a safe, usable value. Ideas are linked logically and use correct terms.

Level 2 (3-4 marks): Covers two strands with some linking, or all three with limited explanation or one error. Reasoning may be partly incomplete, e.g. states current is lower but does not link this to wasted energy.

Level 1 (1-2 marks): Makes one or two relevant statements with little or no linking, e.g. 'step-up transformers increase voltage' or 'high voltage is dangerous in homes'.

Level 0 (0 marks): No response or no creditworthy content.

Indicative content (guidance, not a checklist):
- Step-up transformer at wind farm / power station
- Potential difference increased (e.g. to hundreds of kV)
- Power = potential difference x current, so for same power the current is decreased
- Heating in cables depends on current (P = I^2R); lower current means less energy wasted as thermal energy
- Greater efficiency of transmission / more energy reaches the town
- Step-down transformer(s) near the town
- Potential difference reduced to about 230 V for safe domestic use
- Transformers need alternating current / grid uses a.c.

ALLOW 'voltage' for potential difference. ALLOW 'heat' for thermal energy. ALLOW answers using a worked numerical example if correct.
IGNORE references to resistance of cables being changed by the transformer.
DO NOT ALLOW statements that transformers increase the power or energy, or that a step-up transformer increases current.

(a) \(E = P t = 50 \times 240 = 12\,000\) J.
(b) Some energy is transferred to the surroundings (the block is not insulated), or the thermometer is not in good thermal contact so reads low.
(c) Temperature change = 31 − 19 = 12 °C. \(\Delta Q = m c \Delta\theta = 0.80 \times 900 \times 12 = 8640\) J.
(d) Efficiency = useful energy / total energy = \(8640 / 12\,000 = 0.72\), so 72%.
(e) Insulate the block (lagging) so less energy is lost to the surroundings, so the temperature rise is closer to the true value (more accurate). Repeat the experiment and calculate a mean to reduce the effect of random errors (more reliable). Alternatively use a little water or oil in the thermometer hole for better thermal contact, or wait until the temperature stops rising before reading it.

(a) 12 000 (J) [2]. Correct answer scores 2 marks. If the answer is wrong: substitution of 50 × 240 = 1 mark (method).

(b) Any one from: energy is transferred/lost to the surroundings [1]; thermometer not in good thermal contact with the block [1]. ALLOW 'heat lost to the air'. DO NOT ALLOW 'heat is used up' or 'energy is destroyed'.

(c) 8640 (J) [2]. Correct answer scores 2 marks. If the answer is wrong: temperature change of 12 identified, or substitution 0.80 × 900 × 12 = 1 mark (method). ALLOW 8600 or 8.6 × 10^3.

(d) 72 (%) [2]. Correct answer scores 2 marks. If the answer is wrong: 8640/12 000 or 0.72 = 1 mark (method). ALLOW ecf from (a) and (c) for both marks, provided the answer is a percentage not above 100. ALLOW 0.72 for 1 mark only.

(e) Two changes, each with an explanation, [1] for the change and [1] for the matching explanation. Any two from:
- insulate/lag the block [1] to reduce energy transferred to the surroundings / so temperature rise is more accurate [1];
- put a little oil or water in the thermometer hole [1] for better thermal contact so the thermometer reads the block's true temperature [1];
- repeat the investigation and calculate a mean [1] to reduce the effect of random errors / identify anomalies, making results more reliable [1];
- keep reading the thermometer after switching off until the temperature stops rising [1] so the maximum temperature is recorded [1].
IGNORE 'use a better thermometer' without a reason. Do not award the explanation mark without a valid change.

(c) Power = VI = 6.0 × 0.50 = 3.0 W. Time = 2.0 × 60 = 120 s. Energy = Pt = 3.0 × 120 = 360 J.
(d) p.d. across resistor: \(V = IR = 0.50 \times 8.0 = 4.0\) V. p.d. across lamp = 9.0 − 4.0 = 5.0 V (series circuit, p.d.s add). Resistance of lamp: \(R = V/I = 5.0 / 0.50 = 10\) Ω.

(a) Any four from: ammeter connected in series with the lamp (1); voltmeter connected in parallel (across) the lamp (1); change the potential difference of the supply (or use a variable resistor) in steps (1); record the current and potential difference at each setting (1); at least five different settings / wide range of values (1); repeat readings and calculate a mean / check for anomalies (1). IGNORE switch. DO NOT ALLOW voltmeter in series.
(b) Current increases so the filament gets hotter (1); ions/atoms in the filament vibrate more so there are more collisions with the (flowing) electrons (1). ALLOW 'harder for charge to flow'. DO NOT ALLOW electrons get hotter / lamp gets hotter only.
(c) Correct answer 360 (J) = 3 marks. Otherwise: P = VI = 6.0 × 0.50 or 3.0 W (1); converts 2.0 minutes to 120 s (1); E = Pt = 3.0 × 120 = 360 J (1). ALLOW ecf for the conversion error: 3.0 × 2.0 = 6.0 scores 2 marks.
(d) Correct answer 10 (Ω) = 4 marks. Otherwise: V = IR = 0.50 × 8.0 (1); 4.0 V across resistor (1); 9.0 − 4.0 = 5.0 V across lamp (1); R = V/I = 5.0 / 0.50 = 10 Ω (1). ALLOW ecf from earlier steps. ALLOW total resistance 18 Ω minus 8.0 Ω = 10 Ω (full marks: R = 9.0/0.50 (1), 18 Ω (1), subtract 8.0 (1), 10 Ω (1)).

(a) The ammeter is placed in series with the lamp, the voltmeter is connected in parallel across the lamp only, and the variable resistor is placed in series with the lamp and supply so that the current and potential difference can be changed.

(b) The current increases (but not in proportion to the potential difference).

(c) \(V = I \times R\)

(d) \(R = V \div I = 6.0 \div 0.50 = 12\ \Omega\)

(e) The resistance increases (from 5.0 ohms to 12 ohms).

(f) As the current increases the filament gets hotter, so the ions vibrate more and the flow of electrons is opposed more.

(g) \(P = V \times I = 6.0 \times 0.50 = 3.0\ \text{W}\); \(E = P \times t = 3.0 \times 40 = 120\ \text{J}\)

(a) Any three from: ammeter in series with the lamp [1]; voltmeter in parallel with / across the lamp [1]; variable resistor in series with the lamp [1]; (alternative) vary the current by adjusting the variable resistor / changing the cells [1]. DO NOT ALLOW voltmeter in parallel with the variable resistor or the whole circuit. IGNORE correct battery/switch details.

(b) Increases [1]. ALLOW goes up / gets bigger. DO NOT ALLOW 'directly proportional'.

(c) potential difference = current x resistance, or V = I x R [1]. ALLOW R = V / I or I = V / R. DO NOT ALLOW if symbols have wrong meaning.

(d) Correct answer 12 ohms (or 12 \(\Omega\)) scores [2]. If the answer is wrong or missing: R = V / I or 6.0 / 0.50 [1] (method mark). Unit \(\Omega\) or ohms must be present for the second mark; 12 without unit scores [1]. DO NOT ALLOW 'ohm' misspelt as 'ohms law'.

(e) Increases / gets bigger [1].

(f) The filament / lamp gets hotter (as current increases) [1]. ALLOW higher temperature; ions / atoms vibrate more so more collisions with electrons. IGNORE 'because the current is bigger' on its own.

(g) Correct answer 120 J scores [3]. Otherwise: P = V x I or E = Q x V with Q = I x t [1]; 3.0 W (or Q = 20 C) [1]; 3.0 x 40 (or 20 x 6.0) [1]. ALLOW error carried forward from an incorrect power or charge. ALLOW 1 mark for 3.0 W if no further working.

(a) Primary p.d. \( \times \) primary current = secondary p.d. \( \times \) secondary current.
\( 230 \times I_p = 5.0 \times 2.3 = 11.5 \)
\( I_p = 11.5 \div 230 = 0.05 \) A.

(b) Power transmitted = potential difference \( \times \) current, so for the same power, a higher potential difference means a lower current. Energy wasted as heat in the cables depends on the current (\( P = I^2R \)), so a lower current means much less energy is wasted as heat in the cables.

(a) [3 marks]
Final answer 0.05 A (or 5.0 \( \times 10^{-2} \) A) scores 3 marks.
Otherwise:
- Substitution: \( 230 \times I_p = 5.0 \times 2.3 \) [1]
- Evaluation of right-hand side: 11.5 (W) [1]
- Division: \( 11.5 \div 230 \) giving 0.05 [1]
ALLOW 50 mA. ALLOW 1 mark for 0.05 with wrong or missing unit only if working is shown; DO NOT ALLOW 46 or 0.0217 (wrong rearrangement, e.g. \( 230 \div 5 \) used).

(b) [3 marks]
- For the same power, a higher potential difference means a lower current [1]
- Energy is wasted (as heat) because of the resistance of the cables / the current heats the cables [1]
- Lower current gives less heating / less energy dissipated (so the transmission is more efficient) [1]
ALLOW 'power loss depends on current squared' for the third mark.
IGNORE references to cost or safety on their own.
DO NOT ALLOW 'higher voltage means less resistance' or 'less current is used up'.

Density = mass / volume. A balance measures the mass. The volume of an irregular stone is found from the rise in water level in a measuring cylinder. The answer is C.

C [1]
A, B and D do not allow both mass and volume to be measured.

Timing many swings and dividing reduces the effect of reaction-time error on each single swing. The answer is B.

B [1]
IGNORE references to accuracy only.

For the same power, a higher potential difference means a lower current. A lower current causes less heating in the cables, so less energy is wasted. The answer is D.

D [1]
A is wrong because a step-up transformer reduces current. B describes step-down transformers. C describes a rectifier.

If the live wire touches the case, a large current flows through the earth wire, which melts the fuse and disconnects the appliance. The answer is A.

A [1]
B is the live wire. C is the neutral wire. D is wrong because the earth wire keeps the case at about 0 V.

Total resistance = 4.0 + 6.0 = 10.0 \(\Omega\). \(I = V / R = 5.0 / 10.0 = 0.50\) A. The answer is B.

B [1]
A uses only the 6.0 \(\Omega\) resistor. C uses only the 4.0 \(\Omega\) resistor. D divides the total resistance by the cell voltage (10 / 5).

\(\rho = m / V = 0.24 / 0.000030 = 8000\) kg/m\(^3\). The answer is C.

C [1]
A multiplies mass and volume. B and D invert the equation or mis-handle the powers of ten.

Efficiency = useful output energy / total input energy = 150 / 200 = 0.75, which is 75%. The answer is A.

A [1]
B is the wasted fraction. C and D use wrong ratios; efficiency cannot exceed 100%.

Convert the extension: 3.0 cm = 0.030 m. \(k = F / x = 6.0 / 0.030 = 200\) N/m. The answer is D.

D [1]
C does not convert cm to m. B multiplies force by extension. A inverts the equation.

Power = potential difference × current, so for the same power, a higher potential difference means a lower current. A lower current means less energy is wasted as heat in the cables. The correct answer is C.

C (1 mark). A: current decreases, not increases. B: step-up transformers increase potential difference. D: transformers only work with a.c. and do not change d.c. to a.c.

\(I = V / R = 12 / 4 = 3\ \text{A}\). The correct answer is B.

B (1 mark). A comes from \(R/V\). C comes from \(V - R\). D comes from \(V \times R\).

\(\text{density} = \text{mass} / \text{volume} = 0.60 / 0.00020 = 3000\ \text{kg/m}^3\). The correct answer is D.

D (1 mark). A comes from multiplying mass by volume. B comes from volume ÷ mass. C comes from an error of a factor of 25 in the division.

\(\text{efficiency} = \text{useful output energy} / \text{input energy} = 150 / 200 = 0.75 = 75\%\). The correct answer is C.

C (1 mark). A is the wasted energy as a percentage. B is the difference (200 − 150). D comes from inverting the ratio.

Parallax error happens when the scale is read from an angle. Reading with the eye level with the end of the spring removes this. The correct answer is A.

A (1 mark). B reduces the effect of random errors, not parallax. C and D do not change the viewing angle.

\(v = f \times \lambda = 250 \times 1.4 = 350\ \text{m/s}\). The correct answer is D.

D (1 mark). A comes from \(\lambda / f\). B comes from \(f / \lambda\) (about 179). C is simply the frequency value.

\(F = k x\), so \(x = F / k = 6.0 / 40 = 0.15\ \text{m}\). The correct answer is A.

A (1 mark). B comes from \(k / F\). C comes from \(k - F\). D comes from \(k \times F\).

(a) The National Grid uses alternating current (a.c.).

(b) The UK mains frequency is 50 Hz.

(c) For example solar, tidal, hydroelectric or biofuel.

(d) \( P = VI = 25\,000 \times 80 = 2\,000\,000 \) W.

(e) \( \dfrac{25\,000}{400\,000} = \dfrac{500}{N_s} \), so \( N_s = 500 \times \dfrac{400\,000}{25\,000} = 500 \times 16 = 8000 \) turns.

(f) For a given power, a higher potential difference means a smaller current (\( I = P/V \)). Heating in the cables depends on the current (\( P = I^2R \)), so a smaller current wastes less energy as thermal energy. This makes transmission more efficient.

(g) \( I = P/V = 2\,000\,000 \div 400\,000 = 5 \) A.

(a) a.c. / alternating (current) [1]. DO NOT ALLOW d.c.

(b) 50 Hz / 50 hertz [1]. Both value and unit needed.

(c) Any one from: solar; tidal; hydroelectric; biofuel; geothermal; waves [1]. DO NOT ALLOW fossil fuels or nuclear. IGNORE 'water' alone.

(d) Correct answer 2 000 000 W / 2 × 10^6 W / 2 MW scores 2 marks. Otherwise: substitution 25 000 × 80 [1] (method); evaluation 2 000 000 [1]. ALLOW 1 mark for answer with a power-of-ten error.

(e) Correct answer 8000 scores 2 marks. Otherwise: correct rearrangement or substitution, e.g. Ns = 500 × 400 000 / 25 000 or 25 000/400 000 = 500/Ns [1]; 8000 [1].

(f) Any three linked points [3]:
- power = potential difference × current / for the same power, a higher potential difference means a lower current [1]
- lower current means less heating of the cables (P = I^2R) [1]
- so less energy is wasted / dissipated as thermal energy (to the surroundings) [1]
ALLOW more efficient transmission for the final mark. DO NOT ALLOW 'less energy is lost' without reference to heating or current. DO NOT ALLOW 'the voltage reduces resistance'.

(g) Correct answer 5 A scores 2 marks. Otherwise: I = P / V or 2 000 000 / 400 000 [1]; 5 [1]. ALLOW error carried forward from (d) for the power used.

(a) The meter connected in series measures current: an ammeter.

(b) \(V = IR\), so \(R = \frac{V}{I} = \frac{6.0}{0.50} = 12\ \Omega\).

(c) Time: \(2.0 \times 60 = 120\) s. Power: \(P = VI = 6.0 \times 0.50 = 3.0\) W. Energy: \(E = Pt = 3.0 \times 120 = 360\) J.

(a) ammeter [1]. DO NOT ALLOW voltmeter, multimeter set to volts. IGNORE 'ampmeter' only if clearly intended as ammeter.

(b) Answer 12 (\(\Omega\)) scores 3 marks.
Otherwise: rearrangement \(R = V/I\) [1]; substitution \(6.0/0.50\) [1]; answer 12 [1]. ALLOW 12.0.

(c) Answer 360 (J) scores 3 marks.
Otherwise: converts 2.0 minutes to 120 s [1]; calculates power 3.0 W using \(P = VI\) (or \(E = QV\) route: \(Q = It = 60\) C) [1]; \(E = Pt = 360\) J [1].
ALLOW error carried forward from an incorrect time conversion. ALLOW 3.0 × 2.0 = 6.0 J for 2 marks maximum (no conversion). ALLOW 0.36 kJ.

(a) A higher potential difference means a smaller current for the same power. A smaller current means less energy is transferred as thermal energy in the cables, so the transmission is more efficient.
(b) \(I = P / V = 4\,600\,000 / 230\,000 = 20\) A.
(c) \(P = I^2 R = 20^2 \times 15 = 400 \times 15 = 6000\) W.
(d) 50 Hz.
(e) \(230000 / 11500 = 4000 / N_s\), so \(20 = 4000 / N_s\) and \(N_s = 4000 / 20 = 200\) turns.

(a) [2 marks]
- Higher potential difference means lower current (for the same power) (1)
- so less energy/power is wasted as thermal energy/heating in the cables (1)
ALLOW less energy lost / more efficient for the second mark.
IGNORE 'cheaper' or 'safer' on its own.

(b) [2 marks]
- Correct answer 20 (A) = 2 marks.
- If the answer is incorrect: rearrangement \(I = P/V\) (1) OR correct substitution 4 600 000 / 230 000 (1).
DO NOT ALLOW use of 4.6 without conversion for the accuracy mark.

(c) [2 marks]
- Correct answer 6000 (W) = 2 marks. ALLOW 6 kW.
- If the answer is incorrect: substitution \(20^2 \times 15\) (1).
ALLOW 1 mark for 300 (20 × 15, failure to square).

(d) 50 (Hz) (1)

(e) [2 marks]
- Correct answer 200 (turns) = 2 marks.
- If the answer is incorrect: correct substitution \(230000/11500 = 4000/N_s\) OR ratio 20 seen (1).
ALLOW ecf for a wrong ratio.

(a) A transformer that increases potential difference is a step-up transformer.
(b) The grid voltage is far too high for safe use in homes, so it is stepped down to about 230 V.
(c) \( P = V \times I \), so \( I = \frac{P}{V} = \frac{12\,000\,000}{400\,000} = 30 \) A.
(d) The wasted energy heats the cables and is dissipated to the surroundings as thermal energy.
(e) For a fixed power, a higher potential difference needs a smaller current (\( I = P/V \)). Wasted power is \( I^2R \), so a smaller current gives much less heating in the cables.
(f) 50 Hz.

(a) step-up (transformer) [1]. DO NOT ALLOW 'step-down'.
(b) Any one from: to make the supply safe to use / high potential difference is dangerous / appliances are designed for about 230 V [1]. IGNORE 'to save energy'.
(c) Final answer 30 (A) [2]. If the answer is incorrect, award 1 mark for correct rearrangement \( I = P/V \) or correct substitution \( 12\,000\,000 \div 400\,000 \). ALLOW 3.0 × 10^1 or equivalent for full marks. ALLOW 1 mark for a power-of-ten error in an otherwise correct substitution.
(d) Transferred to the surroundings as thermal energy / heat [1]. ALLOW 'heats the cables / air'. DO NOT ALLOW 'lost' or 'used up' or 'destroyed' on its own.
(e) Any two from: [2]
- higher potential difference gives a lower current (for the same power) [1]
- lower current means less heating (of the cables) / less energy dissipated / smaller \( I^2R \) loss [1]
ALLOW the second mark for 'less energy transferred as thermal energy' only if linked to lower current. IGNORE 'lower resistance'.
(f) 50 (Hz) [1]. Unit required for the mark; ALLOW 'fifty hertz'.

(c) E = P t = 60 × 400 = 24 000 J.
(d) Temperature change = 54 − 20 = 34 °C. Rearranging: c = E ÷ (m × ΔT) = 24 000 ÷ (0.80 × 34) = 24 000 ÷ 27.2 = 882.4... = 880 J/kg °C (2 s.f.).
(e) The same energy gives a larger temperature rise for copper, so copper needs less energy per kg per °C: its specific heat capacity is lower.
(f) Take temperature readings at regular intervals (e.g. every 30 s) while heating, plot temperature against time, and draw a line of best fit. This reduces the effect of a single misread value and anomalies can be spotted; the temperature rise is then read from the line.

(a) thermometer (1). ALLOW temperature probe / temperature sensor / data logger with probe. DO NOT ALLOW thermostat.

(b) Any two from: foam is an insulator / reduces thermal conduction (1); less energy is lost / transferred to the surroundings (1); so more of the heater's energy goes into warming the block / the temperature rise is closer to the true value (1). IGNORE 'keeps the heat in' without reference to energy loss. Max 2.

(c) Final answer 24 000 J (2 marks). Otherwise: correct substitution 60 × 400 (1). ALLOW 2.4 × 10^4. ALLOW 1 mark for 24 with a wrong unit or a power-of-ten error.

(d) Final answer 880 J/kg °C (3 marks). ALLOW 882 or 882.4 for 2 marks if not rounded to 2 s.f. (rounding mark is the third mark).
Otherwise: correct rearrangement c = E ÷ (m ΔT) (1); correct substitution using ΔT = 34 and E = 24 000 (or the candidate's answer from (c), error carried forward) (1); correct evaluation, rounded to 2 s.f. (1). ALLOW 0.88 kJ/kg °C. If ΔT = 54 is used, the substitution mark is lost and max 1 for the rearrangement.

(e) copper's (specific heat capacity) is lower / less (than aluminium) (1). ALLOW less energy needed to raise the temperature of 1 kg by 1 °C. DO NOT ALLOW 'copper heats faster' alone.

(f) Any three from: measure / record the temperature at regular time intervals during heating (1); plot a graph of temperature against time (or energy) (1); draw a line of best fit (1); identify and ignore / repeat anomalous readings (1); use the line / gradient to find the temperature rise or the specific heat capacity (1); repeat the whole investigation and calculate a mean (1). Max 3. IGNORE 'use a better thermometer'.

(a) Rearrange: \( N_s = N_p \times \frac{V_s}{V_p} \).
Substitute: \( N_s = 2200 \times \frac{230}{11000} \).
Evaluate: \( N_s = 46 \).

(b) Power = potential difference x current, so for a given power a higher potential difference means a smaller current. Heating loss in the cables depends on the current (\( P = I^2R \)), so a smaller current wastes much less energy as thermal energy. A step-up transformer at the power station or wind farm raises the potential difference for transmission. A step-down transformer near the town lowers it to a safe value (about 230 V) for homes. This makes the transmission more efficient and reduces cost.

(a) [3 marks]
Correct answer 46 (turns) scores 3 marks.
If the answer is not correct, award:
- 1 mark for correct rearrangement: Ns = Np x Vs / Vp (or equivalent)
- 1 mark for correct substitution: 2200 x 230 / 11000 (or 2200 / (11000/230))
- 1 mark for evaluation: 46
ALLOW 1 mark for substitution of the values into an unrearranged equation (11000/230 = 2200/Ns).
ALLOW 2 marks where the ratio is inverted but the working is otherwise correct and clear (answer 105 000 or similar) - award method marks for rearrangement and substitution only if they are consistent. DO NOT ALLOW answers with Ns = 2200 x 11000 / 230 for the substitution mark.

(b)* Level of response [6 marks]
Level 3 (5-6 marks): Gives a clear, connected explanation that links a higher potential difference to a lower current and reduced energy loss in the cables, AND describes the roles of both step-up and step-down transformers (including the reason for the step-down, safety/domestic use). Logical, well structured, with scientific terms used correctly.
Level 2 (3-4 marks): Describes the roles of both transformers, OR explains the link between high potential difference, low current and reduced energy loss, with some correct reasoning. Some links may be incomplete or missing.
Level 1 (1-2 marks): States one or more relevant simple points (for example high voltage is used, transformers change voltage, less energy is wasted) with little or no explanation.
Level 0 (0 marks): No response or no response worthy of credit.

Indicative scientific points (guidance, not a rigid checklist):
- step-up transformer increases potential difference (after the generator / at the power station or wind farm)
- power = potential difference x current, so for the same power a higher potential difference means a lower current
- lower current means less heating of the cables (wasted thermal energy, P = I^2R)
- so more of the energy reaches the consumers / the transmission is more efficient / cheaper
- step-down transformer(s) decrease potential difference near homes and businesses
- to a safe value (about 230 V) suitable for domestic use
- the National Grid uses alternating current because transformers only work with a.c.
ALLOW 'voltage' for potential difference.
IGNORE references to resistance of cables changing with potential difference.
DO NOT ALLOW 'transformers make electricity' or 'transformers increase power / energy'.
DO NOT ALLOW 'high voltage means less current is used up' (current is not used up).
Best-fit: scientific coverage decides the level; clarity, logical order and substantiation decide the mark within the level.

(a) Power = 6.0 MW = 6 000 000 W. \(I = P \div V = 6\,000\,000 \div 20\,000 = 300\ \text{A}\).

(b) Heating in the cables depends on the current (\(P = I^2R\)). A smaller current means less heating of the cables, so less energy is dissipated to the surroundings and more reaches the consumers.

(c) \(N_s = N_p \times V_s \div V_p = 8000 \times 230 \div 20\,000 = 92\) turns.

(a) [4]
- Converts 6.0 MW to 6 000 000 W (or 6.0 x 10^6 W) [1]
- Correct equation or rearrangement: I = P / V [1]
- Substitution: 6 000 000 / 20 000 [1]
- Answer 300 (A) [1]
If the answer is 300 A award all 4 marks. ALLOW 3.0 x 10^2. If MW is not converted, 0.3 A or 3 x 10^-4 can score max 3 (equation, substitution, evaluation with ecf). DO NOT ALLOW answer without correct unit if a wrong unit is given.

(b) [2]
- Lower current gives less heating (of the cables) / less resistive heating [1]
- So less energy is wasted / dissipated as thermal energy to the surroundings (more energy delivered to consumers) [1]
ALLOW 'power loss depends on current squared' for first mark. IGNORE 'cables get hot' without link to current. DO NOT ALLOW 'less energy is used up / lost' with no mention of heat for the second mark unless heating already stated.

(c) [3]
- Correct equation or rearrangement: Ns = Np x Vs / Vp [1]
- Substitution: 8000 x 230 / 20 000 [1]
- Answer 92 (turns) [1]
If the answer is 92 award all 3 marks. ALLOW ecf for a wrong rearrangement such as 8000 x 20 000 / 230 = 695 652 scoring 1 mark for substitution only.

(a) Power = VI, so for the same power a higher potential difference gives a lower current. A lower current means less heating of the cables (power loss = I^2R), so less energy is wasted.

(b) \( N_p = N_s \times V_p / V_s = 120 \times 11500 / 230 = 120 \times 50 = 6000 \) turns.

(c) \( V_pI_p = V_sI_s \), so \( I_p = (230 \times 40) / 11500 = 9200 / 11500 = 0.80 \) A.

(d) At 0.80 A: \( P = 0.80^2 \times 5.0 = 0.64 \times 5.0 = 3.2 \) W. At 40 A: \( P = 40^2 \times 5.0 = 1600 \times 5.0 = 8000 \) W. 3.2 W is much less than 8000 W, so less power is wasted when the higher potential difference is used.

(a) Any two from, in a linked sequence:
- higher potential difference gives (for the same power) a lower current [1]
- lower current means less heating of the cables / less energy wasted as thermal energy [1]
ALLOW less energy dissipated / lost to the surroundings. IGNORE 'cheaper' or 'safer' on their own. DO NOT ALLOW 'less resistance in the cables'.

(b) Correct answer 6000 (turns) scores 2 marks.
If the answer is incorrect, any one from: rearrangement \( N_p = V_p N_s / V_s \) [1]; substitution \( 11500 \times 120 / 230 \) [1]; \( 11500/230 = 50 \) [1]. Maximum 1 mark if no correct answer.

(c) Correct answer 0.80 (A) scores 2 marks. ALLOW 0.8.
If the answer is incorrect: \( 11500 \times I_p = 230 \times 40 \) or \( I_p = 9200/11500 \) [1].
ALLOW 1 mark for 9200 (W) calculated as the power.

(d) Power loss at 0.80 A = 3.2 W (ALLOW 3.2 W or ecf from (c) using \( 0.80^2 \times 5.0 \)) [1]
Power loss at 40 A = 8000 W [1]
Conclusion that 3.2 W (or ecf) is less than 8000 W, so less power wasted at the higher potential difference [1]
ALLOW ecf for the first mark using the candidate's current from (c). DO NOT ALLOW the conclusion mark without both values being compared.