OxfordAQA International AS and A-level Physics: Wind, solar and hydro energy (9630) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 3.13.2 to 3.13.4 of the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4 (International A-level only). Use g = 9.81 m s⁻² and a water density of 1000 kg m⁻³ unless told otherwise. Air density is given where needed. Full explanations are in the study guide and the revision notes. Course pages: OxfordAQA A-level Physics hub and the checklist. Free diagnostics show which areas need work first.

Questions

1. A wind turbine has blades of length 18.0 m. Air of density 1.2 kg m⁻³ moves at 6.5 m s⁻¹ perpendicular to the rotor. Calculate the maximum power available from the wind. [3]

2. The wind speed at a turbine falls from 12 m s⁻¹ to 9.0 m s⁻¹. (a) Calculate the fraction of the original maximum power now available. [2] (b) At a steady wind speed, the blade length is increased by 30%. State the factor by which the maximum power changes. [2]

3. (a) Explain why the electrical power from a wind turbine is always less than the maximum power available in the wind. [2] (b) Explain why turbines in a wind farm are not placed close together in line with the prevailing wind. [3]

4. Give two advantages and two disadvantages of using a wind farm to supply electricity. [4]

5. The Sun radiates 3.8 × 10²⁶ W. Mars is 2.28 × 10¹¹ m from the Sun. (a) Calculate the intensity of sunlight at Mars. [3] (b) A panel of area 6.0 m² faces the Sun squarely at Mars’s distance and is 15% efficient. Calculate its output. [2]

6. A small lamp radiates equally in all directions. The intensity is 3.2 W m⁻² at 1.5 m. (a) Calculate the power radiated. [2] (b) Calculate the intensity at 4.0 m. [2]

7. A solar cell of area 1.4 cm² is lit by an intensity of 1.0 × 10³ W m⁻². Its readings are:

V / V	0.10	0.30	0.40	0.45	0.50	0.55
I / mA	48	47	45	41	30	0

(a) Calculate the power at each voltage. [2] (b) State the maximum power and the voltage at which it occurs. [1] (c) Calculate the load resistance that gives maximum power. [1] (d) Calculate the efficiency of the cell at maximum power. [2]

8. Each cell in a solar array gives 0.55 V and 90 mA at maximum power. The array must give 4.4 V and 270 mA. State the number of cells in each series branch, the number of parallel branches and the total number of cells, and calculate the power of the array. [4]

9. Describe an experiment with a small lamp and a light-dependent resistor to test whether intensity follows an inverse square law, and how you would use your results to decide whether the law holds. [6]

10. (a) Water falls through a height of 85 m at 22 m³ s⁻¹ in a hydroelectric station that is 82% efficient overall. Calculate the electrical output. [3] (b) A turbine of blade radius 1.8 m stands in a river flowing at 2.5 m s⁻¹. Calculate the maximum power available from the water. [3]

11. (a) Distinguish between a base-power station and a back-up power station. [2] (b) Explain how a pumped storage system works and why it is useful. [3] (c) The upper reservoir holds 1.2 × 10⁶ m³ at a mean height of 250 m above the turbines. Generation is 85% efficient. Calculate the electrical energy it can deliver, in MWh. [3]

Answers

1. P = ½πr²ρv³ [1]; P = ½ × π × 18.0² × 1.2 × 6.5³ [1]; P = 1.68 × 10⁵ W (168 kW) [1] Examiner insight: Show the substitution before the number so your method is visible even if you slip, and give the unit, since leaving it out can weaken an otherwise correct answer.

2. (a) Power ∝ v³, so fraction = (9.0/12)³ [1] = 0.42 [1] (b) Power ∝ r², factor = 1.3² [1] = 1.69 (an increase of 69%) [1] Examiner insight: A ratio method needs the correct power of each variable; writing 0.75 or 1.3 alone is the typical way to lose both marks.

3. (a) The air must leave the turbine still moving, so not all its kinetic energy can be extracted [1]; further energy is lost to friction, turbulence and generator or gearbox heating [1] (b) A turbine produces slower, more turbulent air downwind (a wind shadow) [1]; a turbine in that shadow meets a lower wind speed [1]; power depends on v³, so the loss in output is large, and spacing avoids it [1] Examiner insight: “Loss of energy” alone is too vague for part (a); naming where the energy goes, and linking the shadow to v³ in part (b), is what separates the marks.

4. Advantages, any two: no carbon dioxide emitted when operating; no fuel needed; low running cost [2]. Disadvantages, any two: intermittent output; noise or visual impact; risk to birds and bats; large area needed [2] Examiner insight: Two points in the same bullet, such as “noise” and “sound”, count once; give four distinct ideas.

5. (a) I = P/4πr² [1]; I = 3.8 × 10²⁶ / (4π × (2.28 × 10¹¹)²) [1]; I = 5.82 × 10² W m⁻² [1] (b) Light power = 5.82 × 10² × 6.0 = 3.49 × 10³ W [1]; output = 0.15 × 3.49 × 10³ = 5.2 × 10² W [1] Examiner insight: Squaring the whole bracket 4π × r², not only r, is a frequent slip; carrying the unrounded value into part (b) avoids drift in the last figure.

6. (a) P = 4πr²I = 4π × 1.5² × 3.2 [1] = 90.5 W [1] (b) I = 3.2 × (1.5/4.0)² [1] = 0.45 W m⁻² [1] Examiner insight: Using the ratio in part (b) shows the inverse square law directly; either route should end with the same value, which is a useful check.

7. (a) P = IV: 4.8, 14.1, 18.0, 18.5, 15.0 and 0 mW [2] (b) Maximum 18.5 mW at 0.45 V [1] (c) R = 0.45 / 0.041 = 11 Ω [1] (d) Light power = 1.0 × 10³ × 1.4 × 10⁻⁴ = 0.14 W [1]; efficiency = 0.0185 / 0.14 = 13% [1] Examiner insight: Maximum power must come from computed products, not the largest current or voltage; also convert mA and cm² before comparing powers.

8. Cells per branch = 4.4 / 0.55 = 8 [1]; branches = 270 / 90 = 3 [1]; total = 8 × 3 = 24 cells [1]; power = 4.4 × 0.27 = 1.19 W [1] Examiner insight: State which quantity each arrangement adds; the reasoning is as important as the count.

9. Use a small lamp as the point source in a darkened room, with the LDR on a metre rule in line with it [1]; record the LDR resistance (or conductance) at several distances, repeating each [1]; subtract the reading with the lamp off [1]; measure to the filament and allow for an offset x₀ in the distance, a systematic error [1]; plot 1/√G against d (or check that G × distance² is constant) [1]; if the law holds the graph is straight with a negative d-axis intercept equal to −x₀, within the range tested, because the LDR may not respond exactly in proportion to intensity [1] Examiner insight: Plans that name the variable controlled, the repeat, the background and the graph test tend to gain more credit than ones describing only equipment.

10. (a) P = ρQgh = 1000 × 22 × 9.81 × 85 [1] = 1.83 × 10⁷ W; output = 0.82 × 1.83 × 10⁷ [1] = 1.5 × 10⁷ W (15 MW) [1] (b) P = ½πr²ρv³ [1] = ½ × π × 1.8² × 1000 × 2.5³ [1] = 7.95 × 10⁴ W (about 80 kW) [1] Examiner insight: Part (a) uses a head of water, part (b) a moving stream; choosing the equation that matches the situation is the first mark.

11. (a) A base-power station runs continuously at a steady output to meet constant demand [1]; a back-up station starts quickly to meet extra demand at peak times [1] (b) Water is pumped to a high reservoir when demand is low, using surplus electricity [1]; stored as gravitational potential energy [1]; released through turbines to generate at peak demand, so it supplies back-up power quickly (it is a store, not a source) [1] (c) E = ρVgh = 1000 × 1.2 × 10⁶ × 9.81 × 250 = 2.94 × 10¹² J [1]; = 2.94 × 10¹² / 3.6 × 10⁹ = 817.5 MWh [1]; output = 0.85 × 817.5 = 695 MWh [1] Examiner insight: “Stores energy” with no mention of when it is pumped and when it is released is a common half-answer; include the demand condition.

Where marks are usually lost

• Using diameter, or forgetting the square on r, in the wind power equation.
• Forgetting that v is cubed.
• Giving the maximum power as the power delivered.
• Squaring r but not the factor 4π in the inverse square law.
• Reading the best power from the table without multiplying current by voltage.
• Not converting mA, cm² or J to W, m² or MWh.
• Adding currents in series or voltages in parallel.
• Describing pumped storage as a source of energy.

Next steps

• Revise with the revision notes and the study guide.
• Course pages: OxfordAQA A-level Physics and the checklist.
• Take the free 10-minute diagnostics.
• Next, book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), specification version 4.4, for International AS and A-level exams from May/June 2020, published by OxfordAQA. These questions follow sections 3.13.2, 3.13.3 and 3.13.4.