Secondary 1 Science Semestral Assessment 2 (End of Year) Paper 3

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TuitionGoWhere Practice Paper - Science Secondary 1

TuitionGoWhere Secondary School (AI)

Subject: Science
Level: Secondary 1 (G3)
Paper: SA2 Practice Paper Version 3
Duration: 1 hour 30 minutes
Total Marks: 60

Name: ________________________
Class: ________________________
Date: ________________________

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Instructions to Candidates

1. Write your name, class, and date in the spaces provided above.
2. Answer all questions.
3. Write your answers in the spaces provided on the question paper.
4. The number of marks is given in brackets [ ] at the end of each question or part question.
5. The total number of marks for this paper is 60.
6. You may use a calculator.
7. Where necessary, take the acceleration due to gravity, $g = 10 \text{ N/kg}$.

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Section A: Multiple Choice Questions [15 marks]

Answer all questions. For each question, choose the correct answer and write the letter (A, B, C, or D) in the box provided.

1

A student lifts a 2 kg book from the floor to a shelf 1.5 m high at constant velocity. Which of the following describes the main energy conversion taking place? [1]

☐ A. Kinetic energy → Gravitational potential energy
☐ B. Chemical energy → Gravitational potential energy
☐ C. Chemical energy → Kinetic energy → Gravitational potential energy
☐ D. Gravitational potential energy → Kinetic energy

2

A force of 25 N is used to push a box horizontally across a floor for 4 m. The frictional force acting on the box is 8 N. What is the net work done on the box? [1]

☐ A. 32 J
☐ B. 68 J
☐ C. 100 J
☐ D. 132 J

3

A 500 g ball is thrown vertically upwards with an initial speed of 12 m/s. Ignoring air resistance, what is the maximum height reached by the ball? [1]

☐ A. 3.6 m
☐ B. 7.2 m
☐ C. 14.4 m
☐ D. 28.8 m

4

A simple machine has a mechanical advantage of 4. If an effort of 50 N is applied, what is the maximum load it can lift? [1]

☐ A. 12.5 N
☐ B. 54 N
☐ C. 200 N
☐ D. 400 N

5

The diagram below shows a lever system in equilibrium.

<image_placeholder> id: Q5-fig1 type: diagram linked_question: Q5 description: A uniform lever 1 m long pivoted at its centre. A 20 N weight hangs at the 20 cm mark on the left side. An unknown force F acts downward at the 40 cm mark on the right side. The pivot is at the 50 cm mark. labels: Pivot at centre (50 cm mark), 20 N weight at 20 cm mark (left), Force F at 40 cm mark (right), distances marked from pivot values: Lever length = 1 m, Weight = 20 N at 30 cm from pivot (left), Force F at 10 cm from pivot (right) must_show: Lever arm distances from pivot clearly labelled, force arrows with directions </image_placeholder>

What is the magnitude of force F? [1]

☐ A. 15 N
☐ B. 30 N
☐ C. 60 N
☐ D. 120 N

6

A 60 W lamp is switched on for 30 minutes. How much electrical energy is consumed? [1]

☐ A. 1800 J
☐ B. 108 000 J
☐ C. 180 000 J
☐ D. 1 080 000 J

7

Which of the following statements about power is correct? [1]

☐ A. Power is the total amount of work done.
☐ B. Power is the rate of doing work.
☐ C. Power is measured in joules.
☐ D. Power is the energy stored in an object.

8

A car of mass 1000 kg accelerates from rest to 20 m/s in 10 s. What is the average power developed by the engine? (Assume no energy losses.) [1]

☐ A. 20 000 W
☐ B. 40 000 W
☐ C. 200 000 W
☐ D. 400 000 W

9

An object is dropped from a height of 20 m. At the instant it has fallen 5 m, what is the ratio of its kinetic energy to its gravitational potential energy (relative to the ground)? [1]

☐ A. 1:3
☐ B. 1:4
☐ C. 3:1
☐ D. 4:1

10

A student does 500 J of work in pulling a sled 10 m across horizontal snow. If the force applied is at an angle of 30° to the horizontal, what is the magnitude of the applied force? [1]

☐ A. 28.9 N
☐ B. 50.0 N
☐ C. 57.7 N
☐ D. 100.0 N

11

A block slides down a frictionless inclined plane. Which of the following graphs correctly shows how the kinetic energy (KE) and gravitational potential energy (GPE) of the block change with distance travelled down the plane? [1]

☐ A. KE increases linearly, GPE decreases linearly
☐ B. KE increases exponentially, GPE decreases exponentially
☐ C. KE increases linearly, GPE remains constant
☐ D. KE remains constant, GPE decreases linearly

12

A pulley system has a velocity ratio of 5. An effort of 100 N moves a distance of 2 m to lift a load of 400 N. What is the efficiency of the system? [1]

☐ A. 20%
☐ B. 40%
☐ C. 80%
☐ D. 100%

13

Which energy conversion occurs when a compressed spring is released and pushes a toy car forward along a horizontal floor? [1]

☐ A. Elastic potential energy → Kinetic energy → Heat and sound energy
☐ B. Elastic potential energy → Gravitational potential energy → Kinetic energy
☐ C. Chemical energy → Elastic potential energy → Kinetic energy
☐ D. Kinetic energy → Elastic potential energy → Heat energy

14

A 2 kg object moving at 5 m/s collides with and sticks to a stationary 3 kg object. What is the speed of the combined object after the collision? [1]

☐ A. 1 m/s
☐ B. 2 m/s
☐ C. 3 m/s
☐ D. 5 m/s

15

The diagram shows a velocity-time graph for a 2 kg object.

<image_placeholder> id: Q15-fig1 type: graph linked_question: Q15 description: Velocity-time graph with velocity (m/s) on y-axis and time (s) on x-axis. Graph shows a straight line from (0,0) to (4,8), then horizontal line from (4,8) to (8,8), then straight line down to (12,0). labels: y-axis: Velocity (m/s), x-axis: Time (s), points marked at (0,0), (4,8), (8,8), (12,0) values: Initial acceleration phase 0-4 s, constant velocity 4-8 s, deceleration 8-12 s must_show: Clear axes with units, labelled points, straight line segments </image_placeholder>

What is the total work done on the object during the first 4 seconds? [1]

☐ A. 32 J
☐ B. 64 J
☐ C. 128 J
☐ D. 256 J

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Section B: Structured Questions [30 marks]

Answer all questions in the spaces provided.

16

A crane lifts a concrete block of mass 800 kg vertically upwards at a constant speed of 0.5 m/s through a height of 12 m.

(a) Calculate the weight of the concrete block. [1]

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(b) Calculate the work done by the crane in lifting the block. [2]

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(c) Calculate the power output of the crane. [2]

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(d) The crane's motor has an efficiency of 75%. Calculate the electrical power input to the motor. [2]

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17

A roller coaster car of mass 500 kg starts from rest at point A, which is 40 m above ground level. The track is frictionless. The car passes through point B at a height of 15 m, then goes up to point C at a height of 25 m.

<image_placeholder> id: Q17-fig1 type: diagram linked_question: Q17 description: Roller coaster track profile showing three points: A at 40 m height, B at 15 m height, C at 25 m height. Car shown at point A. Track dips down from A to B then rises to C. labels: Point A (40 m), Point B (15 m), Point C (25 m), ground level (0 m), car at A values: Mass = 500 kg, g = 10 N/kg, heights as labelled must_show: Clear height labels, track shape, car position at A </image_placeholder>

(a) Calculate the gravitational potential energy of the car at point A. [1]

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(b) Calculate the speed of the car at point B. [2]

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(c) Calculate the kinetic energy of the car at point C. [2]

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(d) The actual track has friction. If the car reaches point C with a speed of 12 m/s, calculate the work done against friction between A and C. [2]

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18

A student investigates the relationship between the extension of a spring and the force applied to it. The table shows the results.

Force / N	0	2	4	6	8	10
Extension / cm	0	3	6	9	12	15

(a) Plot a graph of force against extension on the grid below. [2]

<image_placeholder> id: Q18-fig1 type: graph linked_question: Q18 description: Blank graph grid for plotting Force (N) vs Extension (cm). y-axis: Force (N) from 0 to 12, x-axis: Extension (cm) from 0 to 18. labels: y-axis: Force / N, x-axis: Extension / cm, grid lines at 1 cm intervals values: Scale: 1 cm = 1 N on y-axis, 1 cm = 1 cm on x-axis must_show: Labelled axes with units, appropriate scale, grid lines </image_placeholder>

(b) State the relationship between force and extension for this spring. [1]

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(c) Determine the spring constant of the spring in N/m. [2]

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(d) Calculate the elastic potential energy stored in the spring when the force is 10 N. [2]

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19

A block of mass 4 kg is pulled up a rough inclined plane at a constant velocity by a force of 50 N acting parallel to the plane. The plane is inclined at 30° to the horizontal and the block moves a distance of 5 m along the plane.

<image_placeholder> id: Q19-fig1 type: diagram linked_question: Q19 description: Inclined plane at 30° to horizontal. Block of mass 4 kg on plane. Force of 50 N pulling up parallel to plane. Friction force acting down the plane. Weight mg vertically down. Normal reaction perpendicular to plane. labels: 30° angle, 4 kg block, 50 N force up plane, friction force down plane, weight mg, normal reaction, distance 5 m along plane values: Mass = 4 kg, g = 10 N/kg, applied force = 50 N, distance = 5 m, angle = 30° must_show: All forces labelled with arrows, angle marked, distance along plane </image_placeholder>

(a) Calculate the component of the weight acting down the plane. [1]

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(b) Calculate the frictional force acting on the block. [2]

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(c) Calculate the work done against friction. [1]

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(d) Calculate the gain in gravitational potential energy of the block. [2]

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Section C: Longer Structured Questions [15 marks]

Answer all questions in the spaces provided.

20

A hydroelectric power station uses falling water to generate electricity. Water falls through a vertical height of 80 m at a rate of 500 kg/s. The turbine and generator system has an overall efficiency of 85%.

(a) Calculate the gravitational potential energy lost by the water each second. [2]

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(b) Calculate the electrical power output of the power station. [2]

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(c) The electricity generated is transmitted at 110 kV. Calculate the current in the transmission cables. [2]

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(d) Explain why electrical energy is transmitted at high voltage. [2]

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(e) Suggest one environmental advantage and one environmental disadvantage of hydroelectric power. [2]

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(f) During a drought, the water flow rate drops to 300 kg/s while the height remains the same. Calculate the new electrical power output. [2]

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(g) The reservoir behind the dam has a surface area of 2 km². If the water level drops by 0.5 m during the drought, calculate the volume of water lost. [2]

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(h) State the energy conversion that takes place in the turbine. [1]

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End of Paper

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TuitionGoWhere Practice Paper - Science Secondary 1

SA2 Practice Paper Version 3 - Answer Key

Total Marks: 60

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Section A: Multiple Choice Questions [15 marks]

1

Answer: B
Explanation: When a student lifts a book at constant velocity, the chemical energy stored in the student's muscles is converted directly into gravitational potential energy of the book. The kinetic energy does not change (constant velocity), so it is not part of the main energy conversion.
Common mistake: Choosing C because the book moves, but at constant velocity there is no net gain in kinetic energy.

2

Answer: B
Working:
Net force = Applied force - Frictional force = 25 N - 8 N = 17 N
Net work done = Net force × distance = 17 N × 4 m = 68 J
Alternative: Work by applied force = 25 × 4 = 100 J; Work against friction = 8 × 4 = 32 J; Net work = 100 - 32 = 68 J.

3

Answer: B
Working:
Using conservation of energy: Initial KE = Final GPE
$\frac{1}{2}mv^2 = mgh$
$h = \frac{v^2}{2g} = \frac{12^2}{2 \times 10} = \frac{144}{20} = 7.2 \text{ m}$
Mass cancels out, so the 500 g is not needed for the calculation.

4

Answer: C
Working:
Mechanical Advantage = Load / Effort
Load = MA × Effort = 4 × 50 N = 200 N

5

Answer: C
Working:
For equilibrium, sum of clockwise moments = sum of anticlockwise moments about the pivot.
Clockwise moment (from F) = F × 0.10 m
Anticlockwise moment (from 20 N weight) = 20 N × 0.30 m = 6 N·m
F × 0.10 = 6
F = 60 N

6

Answer: D
Working:
Power = 60 W = 60 J/s
Time = 30 minutes = 30 × 60 = 1800 s
Energy = Power × Time = 60 × 1800 = 108 000 J
Wait: 60 × 1800 = 108 000 J. But option D is 1 080 000 J. Let me recalculate.
60 × 1800 = 108 000 J. That's option B.
Correction: Answer is B (108 000 J). The options: A=1800, B=108 000, C=180 000, D=1 080 000.
60 W × 1800 s = 108 000 J. Answer: B

7

Answer: B
Explanation: Power is defined as the rate of doing work, or the rate of energy transfer. Unit: Watt (W) = Joule per second (J/s).
Common mistakes: A confuses power with work; C gives the unit of work/energy; D describes stored energy.

8

Answer: A
Working:
Final KE = $\frac{1}{2}mv^2 = \frac{1}{2} \times 1000 \times 20^2 = 500 \times 400 = 200 000 \text{ J}$
Average power = Work done / Time = 200 000 J / 10 s = 20 000 W

9

Answer: A
Working:
Initial GPE at 20 m = mg × 20
After falling 5 m, height = 15 m
GPE at 15 m = mg × 15
KE gained = Loss in GPE = mg × (20 - 15) = mg × 5
Ratio KE : GPE = (mg × 5) : (mg × 15) = 5 : 15 = 1 : 3

10

Answer: C
Working:
Work done = Force × distance × cos θ
500 = F × 10 × cos 30°
500 = F × 10 × 0.866
500 = F × 8.66
F = 500 / 8.66 ≈ 57.7 N

11

Answer: A
Explanation: On a frictionless inclined plane, GPE decreases linearly with distance (height decreases linearly), and by conservation of energy, KE increases by the same amount, so KE also increases linearly with distance.

12

Answer: B
Working:
Velocity Ratio = Distance moved by effort / Distance moved by load = 5
Distance moved by load = 2 m / 5 = 0.4 m
Work input = Effort × distance moved by effort = 100 N × 2 m = 200 J
Work output = Load × distance moved by load = 400 N × 0.4 m = 160 J
Efficiency = (Work output / Work input) × 100% = (160 / 200) × 100% = 80%
Wait: Let me check. VR = 5, so load moves 1/5 of effort distance. Effort moves 2 m, load moves 0.4 m.
Work in = 100 × 2 = 200 J. Work out = 400 × 0.4 = 160 J. Efficiency = 160/200 = 80%. Answer: C (80%)
Options: A=20%, B=40%, C=80%, D=100%. Answer: C

13

Answer: A
Explanation: The compressed spring stores elastic potential energy. When released, this converts to kinetic energy of the car. As the car moves, friction and air resistance convert kinetic energy to heat and sound energy.

14

Answer: B
Working:
Conservation of momentum (perfectly inelastic collision):
$m_1u_1 + m_2u_2 = (m_1 + m_2)v$
$2 \times 5 + 3 \times 0 = (2 + 3)v$
$10 = 5v$
$v = 2 \text{ m/s}$

15

Answer: B
Working:
First 4 seconds: velocity increases from 0 to 8 m/s uniformly.
Acceleration = (8 - 0) / 4 = 2 m/s²
Force = ma = 2 kg × 2 m/s² = 4 N
Distance in first 4 s = area under graph = $\frac{1}{2} \times 4 \times 8 = 16 \text{ m}$
Work done = Force × distance = 4 N × 16 m = 64 J
Alternative: Work done = Change in KE = $\frac{1}{2}mv^2 - 0 = \frac{1}{2} \times 2 \times 8^2 = 64 \text{ J}$

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Section B: Structured Questions [30 marks]

16

(a) Weight = mg = 800 kg × 10 N/kg = 8000 N [1]

(b) Work done = Force × distance = Weight × height (since constant velocity, force = weight)
= 8000 N × 12 m = 96 000 J [2]
Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

(c) Time taken = distance / speed = 12 m / 0.5 m/s = 24 s
Power = Work done / Time = 96 000 J / 24 s = 4000 W [2]
Mark breakdown: 1 mark for correct time calculation, 1 mark for correct power with unit.

(d) Efficiency = Power output / Power input × 100%
75% = 4000 W / Power input
Power input = 4000 / 0.75 = 5333.33 W (or 5330 W, 5.33 kW) [2]
Mark breakdown: 1 mark for correct rearrangement, 1 mark for correct answer with unit.

17

(a) GPE at A = mgh = 500 kg × 10 N/kg × 40 m = 200 000 J [1]

(b) Loss in GPE from A to B = Gain in KE at B
mg(h_A - h_B) = $\frac{1}{2}mv_B^2$
500 × 10 × (40 - 15) = $\frac{1}{2} \times 500 \times v_B^2$
125 000 = 250 × $v_B^2$
$v_B^2$ = 500
$v_B$ = 22.4 m/s (or $\sqrt{500} = 10\sqrt{5}$ m/s) [2]
Mark breakdown: 1 mark for correct energy conservation equation, 1 mark for correct answer with unit.

(c) At C, height = 25 m. GPE at C = 500 × 10 × 25 = 125 000 J
Total energy = 200 000 J (conserved, frictionless)
KE at C = Total energy - GPE at C = 200 000 - 125 000 = 75 000 J [2]
Mark breakdown: 1 mark for correct GPE at C, 1 mark for correct KE with unit.

(d) Actual KE at C = $\frac{1}{2} \times 500 \times 12^2 = 250 \times 144 = 36 000 \text{ J}$
Expected KE at C (frictionless) = 75 000 J
Work done against friction = Energy lost = 75 000 - 36 000 = 39 000 J [2]
Mark breakdown: 1 mark for correct actual KE, 1 mark for correct work done against friction with unit.

18

(a) Graph plotting: [2]

• Axes labelled with units: Force / N (y-axis), Extension / cm (x-axis) [1]
• All 6 points plotted correctly, straight line through origin [1]
  Expected points: (0,0), (3,2), (6,4), (9,6), (12,8), (15,10)

(b) Force is directly proportional to extension (Hooke's Law is obeyed). [1]
Accept: "Extension is directly proportional to force" or "Linear relationship passing through origin".

(c) Spring constant $k = \frac{F}{x}$ (gradient of graph)
Using any point, e.g., F = 10 N, x = 15 cm = 0.15 m
$k = \frac{10}{0.15} = \mathbf{66.7 \text{ N/m}}$ [2]
Mark breakdown: 1 mark for correct conversion of cm to m, 1 mark for correct calculation with unit.
Alternative: Gradient = (10-0)/(0.15-0) = 66.7 N/m

(d) Elastic potential energy = $\frac{1}{2}Fx = \frac{1}{2} \times 10 \times 0.15 = \mathbf{0.75 \text{ J}}$ [2]
Or using $\frac{1}{2}kx^2 = \frac{1}{2} \times 66.7 \times 0.15^2 = 0.75 \text{ J}$
Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

19

(a) Component of weight down plane = $mg \sin \theta = 4 \times 10 \times \sin 30° = 40 \times 0.5 = \mathbf{20 \text{ N}}$ [1]

(b) Constant velocity → net force = 0
Force up plane = Force down plane
50 N = Friction + Component of weight down plane
50 = Friction + 20
Friction = 30 N [2]
Mark breakdown: 1 mark for stating net force = 0 / forces balanced, 1 mark for correct calculation with unit.

(c) Work done against friction = Friction × distance = 30 N × 5 m = 150 J [1]

(d) Vertical height gained = distance × sin θ = 5 m × sin 30° = 5 × 0.5 = 2.5 m
Gain in GPE = mgh = 4 kg × 10 N/kg × 2.5 m = 100 J [2]
Mark breakdown: 1 mark for correct vertical height, 1 mark for correct GPE with unit.

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Section C: Longer Structured Questions [15 marks]

20

(a) GPE lost per second = mass per second × g × height
= 500 kg/s × 10 N/kg × 80 m = 400 000 J/s (or 400 000 W) [2]
Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

(b) Electrical power output = Efficiency × Power input
= 0.85 × 400 000 W = 340 000 W (or 340 kW) [2]
Mark breakdown: 1 mark for correct efficiency calculation, 1 mark for correct answer with unit.

(c) Power = Voltage × Current
Current = Power / Voltage = 340 000 W / 110 000 V = 3.09 A [2]
Mark breakdown: 1 mark for correct formula/rearrangement, 1 mark for correct answer with unit.

(d) High voltage transmission reduces the current for a given power.
Lower current reduces the power loss in the cables ($P_{loss} = I^2R$), making transmission more efficient. [2]
Mark breakdown: 1 mark for stating reduced current, 1 mark for linking to reduced $I^2R$ losses / increased efficiency.

(e) Advantage: Renewable energy source / no greenhouse gas emissions during operation / no air pollution.
Disadvantage: Disruption of river ecosystems / flooding of land for reservoir / methane from decomposing vegetation / impact on fish migration. [2]
Mark breakdown: 1 mark for valid advantage, 1 mark for valid disadvantage.

(f) New GPE lost per second = 300 kg/s × 10 N/kg × 80 m = 240 000 W
New electrical power output = 0.85 × 240 000 = 204 000 W (or 204 kW) [2]
Mark breakdown: 1 mark for new input power, 1 mark for correct output with unit.

(g) Volume lost = Surface area × drop in water level
= 2 km² × 0.5 m = 2 × (1000 m)² × 0.5 m = 2 × 1 000 000 × 0.5 = 1 000 000 m³ (or $1 \times 10^6 \text{ m}^3$) [2]
Mark breakdown: 1 mark for correct unit conversion (km² to m²), 1 mark for correct calculation with unit.

(h) Kinetic energy of moving water → Mechanical energy (turbine rotation) → Electrical energy
Or: Gravitational potential energy → Kinetic energy → Mechanical energy → Electrical energy [1]
Must show at least three stages with correct sequence.

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Total: 60 marks