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CIE iGCSE Co-Ordinated Science P1.6.3 Energy resources Exam Style Questions Paper 4

CIE iGCSE Co-Ordinated Science Paper 4 - All Chapters

Question

Fig. 3.1 shows a wind turbine used to generate electricity.
(a) State one advantage of generating electricity using wind turbines.
(b) The wind turbine contains an alternating current (a.c.) generator.
On Fig. 3.2, sketch a graph of output voltage against time for the a.c. generator when the wind turbine is turning at a constant speed.
(c) A step-up transformer is used to increase the voltage from the generator.
Describe the construction of a basic step-up transformer.
You may include a labelled diagram to aid your description.
(d) The wind exerts a pressure of \(7200\,\text{Pa}\) on each blade of the wind turbine.
Each blade has a surface area of \(90\,\text{m}^2\).
Calculate the force exerted by the wind on each turbine blade.
(e) The wind turbines produce a low-pitch sound when they turn.
(i) State the minimum frequency of sound which can be heard by a healthy human ear.
(ii) Sound waves are longitudinal waves.
Describe, in terms of oscillations and energy transfer, what is meant by a longitudinal wave.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654, 2025–2027 syllabus):

• Topic P1.6.3 — Energy resources (Part (a))
• Topic P4.5.2 — The a.c. generator (Part (b))
• Topic P4.5.6 — The transformer (Part (c))
• Topic P1.7 — Pressure (Part (d))
• Topic P3.4 — Sound / Topic P3.1 — General properties of waves (Part (e))

▶️ Answer/Explanation

(a) Does not release \(CO_2\) / renewable / no fuel costs

Wind turbines generate electricity without burning fossil fuels.
This means no greenhouse gases are released and the energy source (wind) is renewable.

(b) Sinusoidal waveform

The trace should be a smooth sine wave, alternating above and below the time axis.
Both the amplitude and the time period of the wave must stay constant, since the turbine turns at constant speed.

(c) Soft-iron core with two coils

A basic transformer has a soft iron core linking two separate coils of wire.
The primary and secondary coils are both wound around the same core.
For a step-up transformer, the number of turns on the secondary coil is greater than the number of turns on the primary coil.

(d) \(648\,000\,\text{N}\) (≈ \(6.5\times10^{5}\,\text{N}\))

Force is calculated using \(F = P \times A\).
\(F = 7200 \times 90 = 648\,000\,\text{N}\), which rounds to \(650\,000\,\text{N}\) (2 s.f.).

(e)(i) \(20\,\text{Hz}\)

The typical range of human hearing is approximately \(20\,\text{Hz}\) to \(20\,000\,\text{Hz}\).
\(20\,\text{Hz}\) is therefore the accepted minimum audible frequency for a healthy human ear.

(e)(ii) Oscillations parallel to energy transfer

In a longitudinal wave, the particles oscillate back and forth in the same direction that the energy is being transferred.
This produces alternating regions of compression and rarefaction along the direction of travel.

Question

(a) A simple torch (flashlight) is made from a battery connected to a lamp.
(i) State the name of the energy store in the battery.
(ii) Describe how energy is transferred from this store to the lamp when the lamp is lit.
(b) A ball of mass 0.56 kg is dropped from a height of 9.4 m above the ground.
(i) Calculate the gravitational potential energy transferred by the ball.
(ii) Calculate the speed at which the ball hits the ground.
Air resistance is negligible.
(c) (i) The ball rebounds to a height of 8.2 m.
Air resistance is negligible.
Suggest why the ball does not reach a height of 9.4 m after it bounces.
(ii) Calculate the percentage efficiency of the energy transfer in the bounce.

Most-appropriate topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):

• Topic P1.6.1 — Energy / Energy stores (Part (a)(i))
• Topic P1.6.2 — Work / Energy transfer (Part (a)(ii))
• Topic P1.6.1 — Gravitational potential energy (Part (b)(i))
• Topic P1.6.1 — Kinetic energy / Conservation of energy (Part (b)(ii))
• Topic P1.6.1 — Energy transfers / Energy losses (Part (c)(i))
• Topic P1.6.3 — Efficiency (Part (c)(ii))

▶️ Answer/Explanation

(a)(i) Chemical energy

The battery stores energy in the form of chemical potential energy, which is released during the electrochemical reaction inside the battery.

(a)(ii) Energy is transferred as an electrical current / by electrical work done from the battery to the lamp.

The chemical energy in the battery is converted to electrical energy, which flows through the circuit and is transferred to the lamp, where it becomes light and heat energy.

(b)(i) \( \Delta E_p = mg\Delta h = 0.56 \times 9.8 \times 9.4 = 52 \text{ J} \)

Gravitational potential energy is calculated using mass × gravitational field strength × change in height. The ball loses this energy as it falls to the ground.

(b)(ii) \( E_k = \frac{1}{2}mv^2 \) and \( E_k = E_p = 52 \text{ J} \)

\( 52 = 0.5 \times 0.56 \times v^2 \)
\( v^2 = \frac{52}{0.28} = 185.7 \)
\( v = \sqrt{185.7} = 14 \text{ m/s} \)
Assuming no air resistance, all gravitational potential energy is converted to kinetic energy, allowing us to calculate the impact speed.

(c)(i) Work is done compressing / deforming the ball and by the ball on the ground during the bounce.

Some kinetic energy is transferred to the ground and used to deform the ball, so it is not all converted back to gravitational potential energy. This energy is dissipated as heat and sound.

(c)(ii) Efficiency = \( \frac{\text{useful energy output}}{\text{total energy input}} \times 100 = \frac{8.2}{9.4} \times 100 = 87\% \)

Efficiency is calculated by comparing the height reached after the bounce (useful energy output) to the original drop height (energy input). The ratio of heights gives the efficiency because both heights correspond to gravitational potential energy.
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