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CIE iGCSE Co-Ordinated Science P4.1 Simple phenomena of magnetism Exam Style Questions Paper 4

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Question

Fig. 2.1 shows a person removing a damaged branch from a tree.
(a) The damaged branch has a mass of 225 kg and is lowered 5.2 m to the ground.
Calculate the change in gravitational potential energy (GPE) of the branch as it is lowered to the ground.
The gravitational field strength, \( g = 10 \, \text{N/kg} \).
(b) The damage to the tree was caused by a lightning strike during a thunderstorm.
(i) A scientist estimates that the lightning strike transferred 6000 C of charge in 0.20 s.
Calculate the average current in the lightning strike.
(ii) The thunderstorm produces both light and sound waves.
Explain why an observer sees the light before they hear the sound.
(c) Lightning is caused by electrostatic charges in clouds. Fig. 2.2 shows how charge can form an electric field inside the cloud.
(i) Fig. 2.2 shows negative charge at the base of the cloud.
State the name of the particles that provide this negative charge.
(ii) Describe what is meant by an electric field.
(d) Thunderstorms can produce gamma radiation and X-rays as well as visible light.
Use the phrases to complete the sentences.
You may use each phrase once, more than once or not at all.
less than     more than     the same as
The speed of visible light is ……………………………… the speed of X-rays.
The wavelength of gamma radiation is ……………………………… the wavelength of visible light.
The frequency of X-rays is ……………………………… the frequency of gamma radiation. [2]
(e) When lightning passes through the air, it heats the air up to 10 000°C.
State and explain what happens to the volume of the air when the temperature increases.
Use ideas about molecules in your answer.

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

• Topic P1.6.1 — Energy (Part (a))
• Topic P4.2.2 — Electric current (Part (b)(i))
• Topic P3.4 — Sound (Part (b)(ii))
• Topic P4.1 — Simple phenomena of magnetism (Part (c)(i)–(ii))
• Topic P3.3 — Electromagnetic spectrum (Part (d))
• Topic P2.1.2 — Particle model (Part (e))

▶️ Answer/Explanation

(a) 11 700 J

Change in GPE is calculated using \( \Delta GPE = mgh \).
\( \Delta GPE = 225 \times 10 \times 5.2 = 11\,700 \, \text{J} \).
This energy is transferred to other forms (e.g. kinetic, sound) as the branch falls.

(b)(i) 30 000 A

Current is calculated using \( I = \dfrac{Q}{t} \).
\( I = \dfrac{6000}{0.20} = 30\,000 \, \text{A} \).
This extremely high current reflects the huge charge transfer typical of a lightning strike.

(b)(ii) Light travels faster than sound

Light travels at approximately \( 3 \times 10^8 \, \text{m/s} \), while sound travels at only about \( 340 \, \text{m/s} \) in air.
Both waves travel the same distance from the storm to the observer.
Over that distance, the much slower speed of sound means it arrives noticeably later than the light.

(c)(i) Electrons

Negative charge at the base of a cloud is provided by electrons.
These accumulate due to collisions between ice crystals and water droplets within the cloud.

(c)(ii) A region in which charged particles experience a force

An electric field is defined as a region of space around a charge in which another charged particle experiences a force.
The field lines in Fig. 2.2 point from the positive charge towards the negative charge.

(d) the same as; less than; less than

All electromagnetic waves, including visible light and X-rays, travel at the same speed in a vacuum, \( c = 3 \times 10^8 \, \text{m/s} \).
Gamma radiation has a shorter wavelength than visible light, since it lies further along the electromagnetic spectrum.
Since \( c = f\lambda \) is constant, a shorter wavelength for gamma rays corresponds to a higher frequency, so the frequency of X-rays is less than that of gamma radiation.

(e) Volume increases; molecules move faster and further apart

As temperature increases, air molecules gain more kinetic energy and move faster.
This causes the molecules to spread further apart on average.
As a result, the volume of the air increases (or expands) at constant pressure.

Question

(a) An iron magnet picks up two iron nails as shown in Fig. 6.1.

Explain why the nails do not hang vertically.

▶️Answer/Explanation

Answer: ref. to induced magnetism (in) nails ; two nail heads / north poles / like poles, will repel each other

Detailed Explanation: 1. The magnet induces temporary magnetism in the iron nails 2. Each nail becomes a magnet with: – North pole at end nearest the magnet’s south pole – South pole at end nearest the magnet’s north pole 3. The two adjacent nail heads become like poles (both north or both south) 4. Like magnetic poles repel each other (N-N or S-S repulsion) 5. This causes the nails to splay outward rather than hang straight down

(b) An isotope of iron has a nuclide notation 6026Fe and decays by beta particle emission to an isotope of cobalt.

(i) State what is meant by the term isotope.

▶️Answer/Explanation

Answer: atoms having same atomic number / proton number and different mass number / neutron number

Key Characteristics: 1. Same element (same number of protons) 2. Different numbers of neutrons 3. Same chemical properties (same electron configuration) 4. Different physical properties (density, radioactivity, etc.) 5. Example: 56Fe, 57Fe, and 58Fe are all stable iron isotopes

(ii) Use nuclide notation to complete the symbol equation for this β-decay process.

▶️Answer/Explanation

Answer: 6026Fe → 6027Co + 0-1e

Beta Decay Process: 1. A neutron converts to a proton + electron (β⁻ particle) 2. Atomic number increases by 1 (Fe 26 → Co 27) 3. Mass number remains same (60) 4. The emitted electron is the beta particle (0-1e) 5. Typical half-life for 60Fe is 2.6 million years

(c) An iron wire of length 0.50 m has a cross sectional area of 4.0 × 10-5 m2 and a resistance of 1.21 × 10-3 Ω.
Calculate the resistance of an iron wire of length 0.25 m that has a cross sectional area of 8.0 × 10-5 m2.

▶️Answer/Explanation

Answer: 3.0 × 10-4 Ω

Calculation Steps: Using R = ρL/A (where ρ is resistivity, constant for iron): 1. Original wire: R₁ = ρ(0.50)/(4.0×10-5) = 1.21×10-3 Ω 2. New wire has: – Half the length (0.25 m → decreases resistance) – Double the area (8.0×10-5 m² → decreases resistance) 3. Resistance changes by factor of (½)/(2) = ¼ 4. New resistance R₂ = ¼ × 1.21×10-3 = 3.025×10-4 Ω 5. Round to 2 significant figures: 3.0 × 10-4 Ω

(d) A block of iron is on a bench. The surface of the block of iron in contact with the bench has an area of 144 cm2. The mass of the block of iron is 13.6 kg. Calculate the pressure exerted by the block of iron on the bench in N/cm2. gravitational field strength = 10 N / kg

▶️Answer/Explanation

Answer: 0.94 N/cm2

Calculation: 1. Weight = mass × g = 13.6 kg × 10 N/kg = 136 N 2. Contact area = 144 cm2 3. Pressure = Force/Area = 136 N / 144 cm2 4. = 0.944… N/cm2 5. Round to 2 significant figures: 0.94 N/cm2 Key Points: – Pressure is force per unit area – 1 N/cm2 = 10,000 Pa (SI units) – Iron’s density (7.87 g/cm3) means this block has volume ≈ 1727 cm3

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