CIE iGCSE Co-Ordinated Science P3.4 Sound Exam Style Questions Paper 3
Question


Most-appropriate topic codes (Cambridge IGCSE Coordinated Sciences 0654):
• Topic P1.5.1 — Effects of forces (Part (a)(i), (a)(ii), (a)(iii))
• Topic P3.4 — Sound (Part (b)(i), (b)(ii))
• Topic P4.3.2 — Series and parallel circuits (Part (c)(i), (c)(ii), (c)(iii))
• Topic P3.2.1 — Reflection of light (Part (d))
▶️ Answer/Explanation
(a)(i) D
Force D opposes the motion of the car and is partly caused by air resistance (drag), which increases with speed.
(a)(ii) weight is the gravitational force on an object that has mass / definition of mass (not weight)
The statement describes mass (the quantity of matter). Weight is the gravitational force acting on an object due to its mass in a gravitational field.
(a)(iii) 2000 (N) and (constant speed so) no resultant force
At constant speed, the driving force (B) equals the resisting force (D) because there is no resultant force and no acceleration. Therefore B = D = 2000 N.
(b)(i) decreases
The amplitude of a sound wave is related to loudness. A quieter sound has a smaller amplitude.
(b)(ii) decreases
Pitch is related to frequency. A lower frequency sound has a lower pitch. The engine note becomes lower in pitch.
(c)(i) battery
The battery provides the electromotive force (voltage) that causes current to flow through the circuit.
(c)(ii) 8 (A); current in main part of circuit is greater than current in either branch
In a parallel circuit, the total current from the source equals the sum of the currents in each branch. Current at X = 4 + 4 = 8 A.
(c)(iii) full voltage across both lamps OR if one lamp fails, the other will still work
Parallel connection ensures each lamp receives the full supply voltage and operates independently. If one fails, the other continues to work.
(d) any two from: same size; same distance from mirror; laterally inverted
A plane mirror produces an image that is: same size as the object, same distance behind the mirror as the object is in front, and laterally inverted (left-right reversed). The image is also virtual and upright.
Question

Topic codes (Cambridge IGCSE Co-ordinated Sciences 0654):
• Topic P3.3 — Electromagnetic spectrum
• Topic P3.4 — Sound
• Topic P3.1 — General properties of waves
• Topic P5.2.4 — Half-life
▶️ Answer/Explanation
(a) X-rays are placed between ultraviolet and \(\gamma\)-radiation in the electromagnetic spectrum.
The spectrum is ordered by increasing frequency: radio waves → microwaves → infrared → visible → ultraviolet → X-rays → gamma radiation. X-rays have a higher frequency than ultraviolet light but a lower frequency than gamma radiation, so they belong in the gap between those two.
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(b)(i) One hospital use of X-rays: imaging bones / detecting fractures (radiography).
X-rays penetrate soft tissue but are absorbed by denser materials such as bone, creating shadow images on a detector.
This makes them ideal for diagnosing broken bones, dental problems, or screening for tumours without the need for surgery.
(b)(ii) Ultrasound is used instead of X-rays because it is non-ionising and therefore harmless to the developing foetus.
X-rays are ionising radiation that can damage DNA and cause mutations, posing a serious risk to rapidly dividing foetal cells.
Ultrasound uses high-frequency sound waves that carry no such risk, making it the safe and preferred choice for prenatal scanning.
(b)(iii) Ultrasound has a frequency above 20 kHz — the upper limit of human hearing.
Humans can hear sounds between approximately 20 Hz and 20 kHz; any frequency above this range is called ultrasound.
Medical ultrasound typically operates in the range of thousands of kHz (MHz), well beyond what the human ear can detect.
(c) Wavelength of \(\gamma\)-radiation:
\(\lambda = \dfrac{v}{f} = \dfrac{3 \times 10^{8}}{6 \times 10^{19}} = 5 \times 10^{-12}\,\textbf{m}\)
This extremely short wavelength (5 picometres) is characteristic of gamma radiation and reflects its very high frequency.
All electromagnetic waves share the same wave equation \(v = f\lambda\), with speed \(3 \times 10^8\,\text{m/s}\) in a vacuum.
(d)(i) … half the nuclei of that isotope in any sample to decay.
Half-life is a fixed, characteristic value for each radioactive isotope and is independent of sample size or external conditions.
It describes the statistical rate of random decay — after each half-life, exactly half of the remaining radioactive nuclei will have decayed.
(d)(ii) Number of cobalt-60 atoms remaining after 31.8 years = 25.
\(31.8 \div 5.3 = 6\) half-lives.
\(1600 \times \left(\tfrac{1}{2}\right)^6 = 1600 \div 64 = 25\,\text{atoms}\).
Each half-life halves the count: 1600 → 800 → 400 → 200 → 100 → 50 → 25.
