Question 1
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.3$ — Mass and weight (Part $\mathrm{(a)}$)
• Topic $1.1$ — Physical quantities and measurement techniques (Parts $\mathrm{(b)(i)}$, $\mathrm{(b)(ii)}$)
• Topic $1.7.1$ — Energy (Part $\mathrm{(c)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
$W = 3.0 \text{ N}$
Detailed Solution:
Weight is the gravitational force acting on a mass, given by $W = mg$, where $g = 9.8 \approx 10 \text{ N/kg}$ near Earth’s surface. First convert the mass: $300 \text{ g} = 0.300 \text{ kg}$. Then apply the equation: $W = 0.300 \times 10 = 3.0 \text{ N}$. This result represents the downward gravitational pull on the combined system of masses and hanger. Note that mass is a scalar (quantity of matter), whereas weight is a vector force always directed toward the centre of the Earth. The gravitational field strength $g$ is defined as $g = \dfrac{W}{m}$, confirming that $W = mg$.
(b)(i)
For the correct answer:
$t_{20} = 66.4 \text{ s}$
Detailed Solution:
Reading the stop-watch in Fig. 1.2 directly gives the elapsed time for 20 complete oscillations. The minute hand reads 1 minute and the second hand reads 6.40 s, giving a total of $60 + 6.40 = 66.40 \text{ s}$. This is read to the nearest $0.1 \text{ s}$ (the precision of the stop-watch). Timing 20 oscillations rather than 1 reduces the percentage uncertainty in the period measurement, since any reaction-time error is spread over 20 cycles rather than 1. Always read analogue instruments to the nearest half scale-division.
(b)(ii)
For the correct answer:
$T = 3.32 \text{ s}$
Detailed Solution:
The period $T$ of one oscillation is found by dividing the total time by the number of oscillations: $T = \dfrac{t_{20}}{20} = \dfrac{66.4}{20} = 3.32 \text{ s}$. The period is the time for one complete up-and-down cycle. Timing multiple oscillations and averaging is a standard technique to improve accuracy in experiments involving periodic motion, such as pendulums and mass-spring systems, because it minimises the effect of random errors in starting and stopping the timer.
(c)
For the correct answer (any two):
Elastic potential energy decreases; kinetic energy of masses increases; gravitational potential energy increases.
Detailed Solution:
When the mass hanger is released and moves upward, energy is transferred between stores according to the principle of conservation of energy. The elastic potential energy stored in the stretched band decreases as the band contracts. This energy is transferred to the kinetic energy store of the masses (since they accelerate upward) and simultaneously to the gravitational potential energy store (since $\Delta E_p = mg\Delta h$ increases as height $h$ increases). At the highest point of the oscillation, kinetic energy is momentarily zero and gravitational potential energy is at its maximum. The total mechanical energy ($E_k + E_p$) remains approximately constant, ignoring air resistance and damping.
Question 2
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.5.2$ — Turning effect of forces (Part $\mathrm{(a)}$)
• Topic $1.2$ — Motion (Parts $\mathrm{(b)(i)}$, $\mathrm{(b)(ii)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
Moment $= 45 \text{ N m}$
Detailed Solution:
The moment of a force about a pivot is defined as $\text{moment} = F \times d$, where $F$ is the applied force and $d$ is the perpendicular distance from the pivot to the line of action of the force. Here, $F = 50 \text{ N}$ (the weight of the dumbbell) and $d = 0.90 \text{ m}$ (horizontal distance from shoulder pivot to the weight’s centre of mass). Therefore: $\text{moment} = 50 \times 0.90 = 45 \text{ N m}$. The moment acts in the clockwise direction about the shoulder joint. For the arm to remain stationary (equilibrium), the muscles must supply an equal and opposite anticlockwise moment, illustrating the principle of moments.
(b)(i)
For the correct answer:
Section AB: increasing speed (acceleration); Section DE: stationary (at rest).
Detailed Solution:
On a speed–time graph, the gradient represents acceleration. In section AB, the graph shows an upward slope (increasing speed), meaning the athlete is accelerating — a resultant forward force acts on the athlete. In section DE, the graph is a horizontal line at speed = 0, meaning the athlete is stationary; no net displacement is occurring. A horizontal line at any non-zero speed would indicate constant speed; a horizontal line at zero speed indicates the object is at rest. The area under the graph gives the distance travelled in each section.
(b)(ii)
For the correct answer:
Distance $= 20 \text{ m}$
Detailed Solution:
The distance travelled equals the area under the speed–time graph between $t = 0$ and $t = 5.0 \text{ s}$. From the graph, section AB forms a triangle with base $= 5.0 \text{ s}$ and height $= 8.0 \text{ m/s}$. The area of this triangle gives the distance: $d = \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 5.0 \times 8.0 = 20 \text{ m}$. This method works because for uniformly accelerating motion, the average speed equals half the maximum speed, and $d = \bar{v} \times t = 4.0 \times 5.0 = 20 \text{ m}$, confirming the result.
Question 3
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.4$ — Density (Parts $\mathrm{(a)}$, $\mathrm{(b)(i)}$, $\mathrm{(b)(ii)}$, $\mathrm{(c)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
Volume $= 48 \text{ cm}^3$
Detailed Solution:
The volume of a rectangular block is calculated using $V = l \times b \times h$, where $l$, $b$, and $h$ are the length, breadth, and height respectively. Reading the dimensions from Fig. 3.1 gives (for example) $4.0 \times 3.0 \times 4.0 = 48 \text{ cm}^3$. This formula applies to any cuboid, and it is important to ensure all dimensions are in the same unit before multiplying. Volume is a scalar quantity measured in cm³ or m³. For irregularly shaped solids that sink in liquid, displacement of liquid in a measuring cylinder would instead be used.
(b)(i)
For the correct answer:
$\rho = 11 \text{ g/cm}^3$
Detailed Solution:
Density is defined as mass per unit volume: $\rho = \dfrac{m}{V}$. Substituting the given values: $\rho = \dfrac{86}{8.0} = 10.75 \approx 11 \text{ g/cm}^3$. This high density value is consistent with heavy metals such as lead ($11.3 \text{ g/cm}^3$). Note that $1 \text{ g/cm}^3 = 1000 \text{ kg/m}^3$, so this material has a density of approximately $11\,000 \text{ kg/m}^3$ in SI units. Density is an intrinsic property of a material — it does not change with the size or shape of the sample.
(b)(ii)
For the correct answer:
Any value greater than $11 \text{ g/cm}^3$.
Detailed Solution:
For a solid object to float on a liquid, the density of the liquid must be greater than the density of the solid. This is because an object floats when the upthrust (buoyancy force) equals its weight, which occurs when $\rho_{\text{liquid}} \geq \rho_{\text{solid}}$. Since $\rho_{\text{metal}} = 11 \text{ g/cm}^3$, the liquid must have a density strictly greater than $11 \text{ g/cm}^3$, for example mercury ($\rho \approx 13.6 \text{ g/cm}^3$). If the liquid density equalled the solid’s density, the object would be in neutral buoyancy (submerged but not sinking).
(c)
For the correct answer (any three):
Measure mass of empty measuring cylinder; add liquid and read volume; measure mass of cylinder and liquid; find the difference in the two mass readings.
Detailed Solution:
To determine the density of a liquid using the given equipment: (1) Place the empty measuring cylinder on the balance and record its mass $m_1$. (2) Pour a measured volume $V$ of liquid into the cylinder and read the volume directly from the graduated scale. (3) Reweigh the cylinder with liquid on the balance and record the new mass $m_2$. (4) Calculate the mass of liquid: $m_{\text{liquid}} = m_2 – m_1$. (5) Apply the density equation: $\rho = \dfrac{m_{\text{liquid}}}{V}$. Using a larger volume of liquid reduces the percentage uncertainty in the measurement.
Question 4
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.7.3$ — Energy resources (Parts $\mathrm{(a)}$, $\mathrm{(b)}$, $\mathrm{(c)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
Water flows/falls through pipes to turbines → turbine spins generator → electrical energy produced.
Detailed Solution:
Water stored at height $h$ behind a dam possesses gravitational potential energy given by $\Delta E_p = mg\Delta h$. When released, this energy converts to kinetic energy as water accelerates through pipes (penstocks) toward turbines at the base of the dam. The fast-moving water strikes the turbine blades, causing them to rotate — converting kinetic energy to mechanical (rotational) energy. The spinning turbine shaft drives an electromagnetic generator, which induces an e.m.f. and produces electrical energy by the principle of electromagnetic induction. The overall energy chain is: gravitational potential energy → kinetic energy → electrical energy.
(b)
For the correct answer (any two):
Renewable energy source; no greenhouse gases / CO₂ produced; no SO₂ or acidic gases; no fuel transport needed; output adjustable to meet demand; lakes created for recreation.
Detailed Solution:
Hydroelectric power uses flowing water, which is continuously replenished by the water cycle — making it a fully renewable energy resource, unlike coal which is a finite fossil fuel. During operation, no combustion takes place, so no carbon dioxide, sulfur dioxide, or nitrogen oxides are emitted, eliminating contribution to the greenhouse effect and acid rain. The power output of a hydroelectric station can be rapidly increased or decreased by controlling water flow through the turbines, making it highly responsive to changes in electricity demand — an advantage over many other renewable sources.
(c)
For the correct answer (any two):
Large area of land flooded; damage to wildlife habitats; population displacement; limited suitable sites; changes to downstream water provision; output affected by drought.
Detailed Solution:
Constructing a hydroelectric dam requires flooding large areas of land to form a reservoir, which destroys terrestrial ecosystems and wildlife habitats, and may force human populations to relocate. Suitable sites are geographically limited — large rivers with significant elevation changes are required, making hydroelectric power inaccessible to many regions. The dam also alters the natural flow of the river downstream, affecting water availability for agriculture and other ecosystems. During droughts or periods of low rainfall, the reservoir level falls, potentially reducing or halting electricity generation entirely.
Question 5
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.5.1$ — Effects of forces (Part $\mathrm{(a)}$)
• Topic $1.8$ — Pressure (Part $\mathrm{(b)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
Resultant force $= 30 \text{ N}$, direction: to the left (forwards).
Detailed Solution:
The resultant force is found by adding all forces along the same line, taking direction into account. From Fig. 5.1, the pushing force is $120 \text{ N}$ (to the left/forwards) and the friction force is $90 \text{ N}$ (opposing motion, to the right). The resultant is: $F_{\text{net}} = 120 – 90 = 30 \text{ N}$ in the direction of the push (to the left/forwards). Since $F_{\text{net}} \neq 0$, the trolley accelerates in the direction of the push. By Newton’s second law, $F = ma$, this net force causes acceleration proportional to its magnitude and in its direction.
(b)
For the correct answer:
Pressure $= 28 \text{ N/cm}^2$
Detailed Solution:
Pressure is defined as force per unit area: $p = \dfrac{F}{A}$. The total contact area of all four wheels is $A = 4 \times 8.0 = 32 \text{ cm}^2$. The force pressing down on the ground equals the total weight: $F = 900 \text{ N}$. Substituting: $p = \dfrac{900}{32} = 28.125 \approx 28 \text{ N/cm}^2$. The unit N/cm² is appropriate here since area was given in cm². Converting to SI: $28 \text{ N/cm}^2 = 28 \times 10^4 \text{ Pa} = 280\,000 \text{ Pa}$. Pressure increases if the contact area decreases or the force increases — this principle explains why sharp objects penetrate surfaces more easily.
Question 6
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $2.1.2$ — Particle model (Parts $\mathrm{(a)}$, $\mathrm{(b)}$)
• Topic $2.1.3$ — Gases and the absolute scale of temperature (Part $\mathrm{(c)}$)
▶️ Answer/Explanation
(a)
For the correct answer (any two):
Move at high speed; in random directions; with collisions between molecules and walls.
Detailed Solution:
According to the kinetic particle model, gas particles are in continuous, rapid, random motion. They travel in straight lines between collisions, changing direction only when they collide with other particles or with the container walls. The motion is described as random because there is no preferred direction — particles move with a range of speeds in all directions. At room temperature, average molecular speeds in air are approximately $500 \text{ m/s}$. The random nature of the motion is evidenced by Brownian motion, where microscopic particles in suspension are seen to move erratically due to unequal bombardment by surrounding gas molecules.
(b)
For the correct answer:
Particles move more slowly (average speed decreases).
Detailed Solution:
Temperature is a measure of the average kinetic energy of the particles in a substance. When the gas is cooled, thermal energy is removed, so the average kinetic energy of the particles decreases. Since kinetic energy $E_k = \dfrac{1}{2}mv^2$, a decrease in $E_k$ means a decrease in the average speed $v$ of the particles. At absolute zero ($-273°C$ or $0 \text{ K}$), the particles would have minimum kinetic energy. Cooling also reduces the rate and force of collisions with the container walls, which explains why gas pressure decreases at lower temperatures (at constant volume).
(c)
For the correct answer:
Particles collide with walls; collisions apply a force to the walls; force spread over area creates pressure.
Detailed Solution:
Gas particles move rapidly and continuously collide with the walls of the container. Each collision exerts a small force on the wall as the particle changes momentum (by Newton’s second law, force equals rate of change of momentum: $F = \dfrac{\Delta p}{\Delta t}$). Although each individual collision exerts a tiny force, the enormous number of collisions per second — billions per square centimetre — produces a large, effectively constant, total force on the wall. Pressure is this total force divided by the wall area: $p = \dfrac{F}{A}$. Increasing temperature increases particle speed and collision frequency, thereby increasing pressure.
Question 7
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $2.3.1$ — Conduction (Part $\mathrm{(a)}$)
• Topic $2.3.2$ — Convection (Part $\mathrm{(b)}$)
• Topic $2.3.3$ — Radiation (Part $\mathrm{(c)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
Conduction
Detailed Solution:
In metals, thermal energy is transferred by two mechanisms: lattice vibrations and — more importantly — the movement of free (delocalised) electrons. When one end of a metal is heated, the free electrons in that region gain kinetic energy and move rapidly throughout the metal lattice, colliding with other electrons and ions and transferring energy along the conductor. This is why metals are far better thermal conductors than non-metals, which lack free electrons. Conduction requires direct contact between particles and does not require bulk movement of the material itself.
(b)
For the correct answer (any three):
Convection; heated water expands; becomes less dense; less dense water rises / denser water sinks.
Detailed Solution:
The process shown in Fig. 7.1 is convection, the dominant method of thermal energy transfer in liquids. When the base of the can is heated, the water near the bottom gains thermal energy, causing its particles to move faster and further apart — the water expands and its density decreases. Since this less dense, warmer water is surrounded by cooler, denser water above, buoyancy forces push the warm water upward. The cooler, denser water descends to replace it near the heat source, forming a circular convection current. This current continuously transfers thermal energy throughout the entire volume of water.
(c)
For the correct answer:
The black can has the greater temperature increase; black surfaces are better absorbers of infrared radiation than white surfaces.
Detailed Solution:
Thermal radiation from the Sun is primarily infrared radiation. The ability of a surface to absorb radiation depends on its colour and texture: dull, dark (black) surfaces are the best absorbers of infrared radiation, while shiny, light (white) surfaces are poor absorbers and good reflectors. Therefore, the black can absorbs more infrared radiation per unit time than the white can, transferring more thermal energy to the water inside it. Since both cans contain equal masses of water with the same specific heat capacity $c$, the greater energy input into the black can ($Q = mc\Delta\theta$) results in a larger temperature rise $\Delta\theta$.
Question 8
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $3.1$ — General properties of waves (Parts $\mathrm{(a)(i)}$, $\mathrm{(a)(ii)}$, $\mathrm{(b)(i)}$, $\mathrm{(b)(ii)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
One amplitude correctly drawn and labelled A on Fig. 8.1.
Detailed Solution:
The amplitude $A$ of a wave is the maximum displacement of a point on the wave from its equilibrium (undisturbed) position. On the displacement–distance graph, it is the vertical distance from the midline (zero displacement) to either a crest (maximum positive displacement) or a trough (maximum negative displacement). It should be marked as a vertical double-headed arrow from the centre line to either the peak or trough. Note that amplitude is not the distance from crest to trough — that would be $2A$. Amplitude is related to the energy carried by the wave: $E \propto A^2$.
(a)(ii)
For the correct answer:
One wavelength correctly drawn and labelled L on Fig. 8.1.
Detailed Solution:
The wavelength $\lambda$ (labelled L here) is the shortest distance between two points on the wave that are in phase — i.e., at the same position in their oscillation cycle with the same displacement and velocity. On a displacement–distance graph, it can be measured from crest to crest, trough to trough, or between any two corresponding points separated by one complete cycle. It should be shown as a horizontal double-headed arrow along the direction of wave propagation. Wavelength is related to wave speed and frequency by $v = f\lambda$.
(b)(i)
For the correct answer:
Three reflected wavefronts drawn at equal spacing; direction arrow pointing vertically downward (perpendicular to reflected wavefronts).
Detailed Solution:
When plane wavefronts strike a flat barrier at 45°, they obey the law of reflection: the angle of incidence equals the angle of reflection ($i = r$), both measured from the normal to the barrier. With the barrier at 45° to the incoming wavefronts, the reflected wavefronts travel perpendicular to the incident wavefronts — in this case, vertically downward in the diagram. The reflected wavefronts must be drawn parallel to each other, with the same spacing (wavelength) as the incident wavefronts, since reflection does not change wave speed or frequency. The direction arrow must be perpendicular to the reflected wavefronts.
(b)(ii)
For the correct answer:
Three semicircular wavefronts centred on the gap; same wavelength (spacing) as the incident wavefronts.
Detailed Solution:
When plane waves pass through a narrow gap (where the gap width is comparable to or smaller than the wavelength), they undergo diffraction — they spread out into the region beyond the gap. The diffracted wavefronts are semicircular, centred on the gap, spreading in all directions beyond the barrier. The wavelength (spacing between successive wavefronts) remains unchanged because diffraction does not alter wave speed or frequency. Greater diffraction occurs when the gap width is smaller relative to the wavelength. This phenomenon is described by Huygens’ principle, which treats every point on a wavefront as a new source of secondary wavelets.
Question 9
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $3.3$ — Electromagnetic spectrum (Parts $\mathrm{(a)(i)}$, $\mathrm{(a)(ii)}$, $\mathrm{(b)}$, $\mathrm{(c)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
Microwaves
Detailed Solution:
The electromagnetic spectrum, in order of increasing frequency (decreasing wavelength), runs: radio waves → microwaves → infrared → visible light → ultraviolet → X-rays → gamma rays. From Fig. 9.1, the unlabelled region lies between radio waves and infrared, which corresponds to microwaves. Microwaves have wavelengths ranging from approximately $1 \text{ mm}$ to $1 \text{ m}$ and frequencies from about $300 \text{ MHz}$ to $300 \text{ GHz}$. They are used in satellite television, mobile phone networks, and microwave ovens. All electromagnetic waves travel at the same speed in a vacuum: $c = 3.0 \times 10^8 \text{ m/s}$.
(a)(ii)
For the correct answer:
Frequency
Detailed Solution:
The arrow in Fig. 9.1 points from radio waves toward gamma rays, which is the direction of increasing frequency. This is consistent with the relationship $c = f\lambda$: since all EM waves travel at the same speed $c$ in a vacuum, higher frequency corresponds to shorter wavelength. Gamma rays have the highest frequency and shortest wavelength, while radio waves have the lowest frequency and longest wavelength. Alternatively, the arrow could be described as showing increasing photon energy ($E = hf$) or decreasing wavelength, but the mark scheme specifies frequency.
(b)
For the correct answer (any two):
Medical imaging/scanning of bones and internal organs; treating cancer; bone density measurement; security scanning of baggage; space telescopes.
Detailed Solution:
X-rays have short wavelengths ($10^{-8}$ to $10^{-13}$ m) and high penetrating power through soft tissue but are absorbed by denser materials such as bone and metal. In medicine, X-rays are used to produce shadow images of bones to detect fractures, and in CT scans to produce detailed cross-sectional images of internal organs. High-energy X-rays are also used in radiotherapy to destroy cancerous tumours. In security, X-ray scanners at airports reveal concealed objects in luggage without opening bags. Each application exploits the penetrating nature of X-rays combined with differential absorption in different materials.
(c)
For the correct answer (any two):
X-rays are ionising radiation; they damage body cells/tissue/DNA; they can cause mutations or cancer.
Detailed Solution:
X-rays are ionising radiation, meaning they carry sufficient energy to remove electrons from atoms, creating ions. When X-rays pass through living tissue, they ionise molecules within cells, particularly DNA. This ionisation can break chemical bonds in the DNA double helix, causing mutations — random changes to the genetic code. If the DNA damage is not repaired correctly, affected cells may undergo uncontrolled division, leading to cancer. Higher doses increase the probability of harm, which is why medical X-ray exposure is kept as low as reasonably achievable (ALARA principle), and radiographers stand behind lead screens for protection.
Question 10
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $4.2.4$ — Resistance (Parts $\mathrm{(a)(i)}$, $\mathrm{(a)(ii)}$)
• Topic $4.3.2$ — Series and parallel circuits (Part $\mathrm{(b)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
Correct voltmeter symbol (circle with V); connected in parallel with the LDR.
Detailed Solution:
A voltmeter measures the potential difference (voltage) across a component and must always be connected in parallel with that component — i.e., with one terminal connected to each end of the LDR. The voltmeter symbol is a circle containing the letter V. An ideal voltmeter has infinite resistance so that it draws negligible current from the circuit and does not affect the measurement. In contrast, an ammeter is connected in series and has (ideally) zero resistance. Connecting a voltmeter in series would be incorrect as it would have a very high resistance and prevent current from flowing through the circuit.
(a)(ii)
For the correct answer:
$R = 270 \text{ Ω}$
Detailed Solution:
Resistance is defined by Ohm’s Law (for a resistor of constant resistance): $R = \dfrac{V}{I}$. Substituting the given values — potential difference $V = 5.4 \text{ V}$ and current $I = 0.020 \text{ A}$: $R = \dfrac{5.4}{0.020} = 270 \text{ Ω}$. An LDR (light-dependent resistor) is a component whose resistance decreases as the intensity of light falling on it increases, due to photons freeing more charge carriers in the semiconductor material. At low light levels, the resistance is very high (many kΩ); at high light levels it drops to a few hundred ohms, as seen here.
(b)
For the correct answer:
Combined resistance $= 330 \text{ Ω}$
Detailed Solution:
When resistors are connected in series, the combined (total) resistance is the sum of the individual resistances: $R_{\text{total}} = R_1 + R_2 + \ldots$ The LDR and variable resistor are connected in series in this circuit, so: $R_{\text{total}} = 150 + 180 = 330 \text{ Ω}$. In a series circuit, the same current flows through all components, and the total resistance increases with each additional component added. This is in contrast to a parallel arrangement, where the combined resistance is always less than the smallest individual resistance, calculated using $\dfrac{1}{R_{\text{total}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$.
Question 11
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $4.5.6$ — The transformer (Part $\mathrm{(a)}$)
• Topic $4.4$ — Electrical safety (Part $\mathrm{(b)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
$V_s = 20 \text{ V}$
Detailed Solution:
The turns ratio equation for an ideal transformer relates primary and secondary voltages to the number of coil turns: $\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}$. Rearranging for the secondary voltage: $V_s = V_p \times \dfrac{N_s}{N_p} = 240 \times \dfrac{48}{590} = 240 \times 0.0814 \approx 19.5 \approx 20 \text{ V}$. Since $N_s < N_p$, this is a step-down transformer — it reduces the voltage from 240 V to approximately 20 V, which is suitable for charging a laptop battery. The ratio $N_p : N_s = 590 : 48 \approx 12.3 : 1$ is the turns ratio, equal to the voltage ratio for an ideal (100% efficient) transformer.
(b)
For the correct answer:
A large (excess) current flows through the fuse; the fuse wire heats up and melts; this breaks the circuit, isolating the appliance from the supply.
Detailed Solution:
A fuse is a safety device consisting of a thin wire of low melting point (typically tin-lead alloy) placed in series with the live wire of a circuit. Under normal operating conditions, the current is within the rated value and the fuse wire carries it safely. If a fault develops — such as a short circuit or overload — an excessively large current flows. This large current generates heat in the fuse wire by the joule heating effect ($P = I^2R$), raising its temperature above its melting point. The wire melts (blows), permanently breaking the circuit and cutting off current to the appliance, preventing fire or electric shock. A blown fuse must be replaced with one of the correct rating, never bypassed.
Question 12
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $5.2.2$ — The three types of nuclear emission (Parts $\mathrm{(a)(i)}$, $\mathrm{(a)(ii)}$)
• Topic $5.1.2$ — The nucleus (Part $\mathrm{(b)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
1st empty box (beta particle type): electron; 2nd empty box (alpha charge): $+2$; 3rd empty box (gamma ionising ability): high.
Detailed Solution:
Alpha ($\alpha$) particles consist of 2 protons and 2 neutrons (equivalent to a helium-4 nucleus, $^4_2\text{He}$), carrying a charge of $+2$. Beta ($\beta^-$) particles are fast-moving electrons ($^0_{-1}e$) emitted from the nucleus when a neutron decays into a proton. Gamma ($\gamma$) radiation consists of high-energy electromagnetic photons with no charge and no mass, and has the highest penetrating power of the three. Alpha particles are the most strongly ionising (due to their large charge and mass) but the least penetrating; gamma rays are the least ionising but the most penetrating, requiring several centimetres of lead or metres of concrete for significant absorption.
(a)(ii)
For the correct answer:
Gamma ($\gamma$)
Detailed Solution:
Gamma radiation is the most penetrating of the three types of nuclear emission. Alpha particles are stopped by a few centimetres of air or a thin sheet of paper. Beta particles are stopped by a few millimetres of aluminium. Gamma rays, being high-energy electromagnetic radiation with no charge and no mass, can penetrate several centimetres of dense lead or many metres of concrete before being significantly attenuated. Penetrating power is inversely related to ionising ability: gamma rays ionise weakly (fewer interactions per unit path length) but travel very far, while alpha particles ionise intensely but are stopped very quickly. This is why gamma sources require the most shielding for radiation safety.
(b)
For the correct answer:
X = proton(s); Y = neutron(s)
Detailed Solution:
The nuclide notation $^{9}_{4}\text{Be}$ tells us that beryllium-9 has a proton number (atomic number) $Z = 4$ and a nucleon number (mass number) $A = 9$. The nucleus therefore contains $Z = 4$ protons and $A – Z = 9 – 4 = 5$ neutrons. In Fig. 12.1, the particles labelled X inside the nucleus that carry a positive charge ($+1$ each) are the protons, while the neutral particles labelled Y are the neutrons. The 4 electrons (equal to the number of protons) orbit the nucleus in shells, keeping the atom electrically neutral overall. Isotopes of beryllium share the same proton number (4) but differ in neutron number.