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CIE iGCSE Co-Ordinated Science P1.7 Pressure Exam Style Questions Paper 3

Question

(a) The Sun is the closest star to the Earth.
(i) State the name of the force that keeps the Earth in orbit around the Sun.
(ii) State the name of the galaxy which contains the Sun.
(b) Energy from the Sun is used to power an electric car.
The energy is stored in a battery in the car.
The battery supplies a current of 96 A at 120 V to the motor that drives the car.
(i) Calculate the electrical energy transferred to the motor in 900 s. State the unit of your answer.
(ii) The current supplied by the battery is direct current (d.c.). Describe the difference between direct current (d.c.) and alternating current (a.c.).
(c) The car driver uses a wrench to remove a wheel from the car. The driver puts the wrench on a wheel nut as shown in Fig. 10.1.
(i) The driver uses a force of 300 N at a distance 0.40 m from the wheel nut. Calculate the moment of the force about the centre of the wheel nut.
(ii) Fig. 10.2 shows the tyre in contact with the road.
The area in contact with the road is 180 cm².
The tyre exerts a pressure of 20 N/cm² on the road.
Calculate the force exerted by the tyre on the road.

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

• Topic P6.1 — The Solar System (Part (a))
• Topic P4.2.5 — Electrical energy and electrical power (Part (b)(i))
• Topic P4.2.2 — Electric current (Part (b)(ii))
• Topic P1.5.2 — Turning effect of forces (Part (c)(i))
• Topic P1.7 — Pressure (Part (c)(ii))

▶️ Answer/Explanation

(a)(i) Gravitational attraction
The Sun’s gravitational field provides the centripetal force that keeps the Earth in its nearly circular orbit around the Sun.

(a)(ii) Milky Way
The Sun is one of approximately 100–400 billion stars in the Milky Way galaxy, which is a spiral galaxy about 100,000 light-years in diameter.

(b)(i) Energy = \(I \times V \times t = 96 \times 120 \times 900 = 10,368,000 \text{ J} \approx 10,000,000 \text{ J}\)
The electrical energy transferred is calculated using the formula \(E = IVt\), where current is in amperes, voltage in volts, and time in seconds. The unit of energy is the joule (J).

(b)(ii) Direct current (d.c.) flows in one direction only, whereas alternating current (a.c.) periodically changes direction.
In a direct current circuit, electrons flow continuously in the same direction. In an alternating current circuit, the direction of electron flow reverses periodically (e.g., 50 times per second in mains electricity).

(c)(i) moment = \(300 \times 0.40 = 120 \text{ N m}\)
The moment of a force is calculated using the equation \(\text{moment} = \text{force} \times \text{perpendicular distance from the pivot}\). The unit is newton-metre (N m).

(c)(ii) force = \(20 \times 180 = 3600 \text{ N}\)
Pressure is defined as force per unit area. The force can be calculated by multiplying pressure by area: \(\text{force} = \text{pressure} \times \text{area}\).

Question

(a) Fig. 11.1 shows an elephant pushing a tree trunk along at a constant speed. The elephant exerts a constant force of 1500 N to move the tree trunk 20 m in the direction of the force.
(i) Calculate the work done by the elephant when the tree trunk is moved 20 m.
(ii) The elephant stands with all four feet on the ground. The area of each foot in contact with the ground is \(0.070 \, \text{m}^2\). The weight of the elephant is \(36,000 \, \text{N}\). Calculate the pressure exerted on the ground due to the elephant.
(b) Table 11.1 shows the highest- and lowest-frequency sounds that four animals are able to hear.
(i) State which animal in Table 11.1 has the smallest audible frequency range.
(ii) State the approximate range of frequencies audible to humans.

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

• Topic P1.6.2 — Work (Part (a)(i))
• Topic P1.7 — Pressure (Part (a)(ii))
• Topic P3.4 — Sound (Part (b))

▶️ Answer/Explanation

(a)(i) 30,000 J

Work done = force × distance moved in the direction of the force.
Work done = \(1500 \, \text{N} \times 20 \, \text{m} = 30,000 \, \text{J}\) (or \(3.0 \times 10^4 \, \text{J}\)).
The unit of work is the joule (J).

(a)(ii) 130,000 N/m²

Pressure = force ÷ area.
Total area in contact with the ground = \(4 \times 0.070 = 0.28 \, \text{m}^2\).
Pressure = \(\frac{36,000}{0.28} = 128,571 \, \text{N/m}^2 \approx 130,000 \, \text{N/m}^2\) (or 130,000 Pa).

(b)(i) elephant

Frequency range = highest frequency − lowest frequency:

  • Bat: 110,000 − 2,000 = 108,000 Hz
  • Dog: 50,000 − 50 = 49,950 Hz
  • Elephant: 12,000 − 5 = 11,995 Hz (smallest range)
  • Mouse: 100,000 − 1,000 = 99,000 Hz

(b)(ii) 20 Hz to 20,000 Hz

The typical human audible frequency range is approximately 20 Hz (lowest frequency heard) to 20,000 Hz (highest frequency heard). This range varies with age and individual hearing ability.

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