Question 1
State the two measurements.
For each measurement, suggest the instrument used for making the measurement.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.1$ — Physical quantities and measurement techniques (Part $\mathrm{(a)}$)
• Topic $1.2$ — Motion (Parts $\mathrm{(b)(i)}$, $\mathrm{(b)(ii)}$, $\mathrm{(b)(iii)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
Measurement 1: distance, Instrument: metre rule(r)
Measurement 2: time, Instrument: stopwatch
To find the average speed, we use the core physics formula $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$. Therefore, the two measurements required are the distance the trolley travels down the slope and the time it takes to cover that exact distance. You can easily use a standard metre rule to measure the physical distance and a stopwatch to reliably measure the elapsed time.
(b)(i)
For the correct answer:
$12.5\text{ cm/s}$
Looking at the speed-time graph in Fig 1.1, simply locate the $2.0\text{ s}$ mark on the x-axis (time). By moving vertically up to the plotted line and horizontally across to the y-axis (speed), you can read the corresponding speed value exactly midway between 10 and 15, which is $12.5\text{ cm/s}$.
(b)(ii)
For the correct answer:
$50\text{ cm}$
The total distance travelled is represented by the area underneath the speed-time graph. For the first $4.0\text{ s}$, this area forms a right-angled triangle. So, $\text{Distance} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 25\text{ cm/s} = 50\text{ cm}$.
(b)(iii)
For the correct answer:
From $0$ to $4.0\text{ s}$: constant acceleration.
From $4.0\text{ s}$ to $10\text{ s}$: constant speed.
From $0\text{ s}$ to $4.0\text{ s}$, the graph shows a straight, upward-sloping line, which indicates that the trolley is gaining speed at a steady rate (constant acceleration). From $4.0\text{ s}$ to $10\text{ s}$, the line levels out and becomes completely horizontal, meaning the speed is no longer changing and the trolley continues at a constant speed.
Question 2
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.1$ — Physical quantities and measurement techniques (Part $\mathrm{(a)}$)
• Topic $1.3$ — Mass and weight (Part $\mathrm{(b)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
$0.11\text{ mm}$
To find the average thickness of a single sheet, you just need to divide the total thickness of the entire stack of paper by the total number of sheets. By dividing $29\text{ mm}$ by $270$, you get roughly $0.107\text{ mm}$, rounding neatly to $0.11\text{ mm}$ per sheet. This method is incredibly helpful because it effectively averages out any microscopic differences between individual sheets.
(b)
For the correct answer:
$13\text{ N}$
Weight is simply the force of gravity acting on an object’s mass and is calculated using the formula $W = mg$, where $m$ is the mass in kilograms and $g$ is the gravitational field strength (taken as $10\text{ N/kg}$ near Earth’s surface in this syllabus). First, you must convert the mass from grams to kilograms ($1300\text{ g} = 1.3\text{ kg}$). Multiplying $1.3\text{ kg}$ by $10$ gives a total weight of $13\text{ N}$.
Question 3
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.5.1$ — Effects of forces (Parts $\mathrm{(a)(i)}$, $\mathrm{(a)(ii)}$, $\mathrm{(a)(iii)}$)
• Topic $1.5.2$ — Turning effect of forces (Part $\mathrm{(b)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
Resultant force = $4000\text{ N}$
Direction = forwards (to the right)
The resultant force is the overall net force acting on the aeroplane once all opposing forces are combined. Since the forward thrust ($12000\text{ N}$) and the backward resistance ($8000\text{ N}$) act in completely opposite directions, you simply subtract the smaller force from the larger one, leaving an unbalanced force of $4000\text{ N}$ pushing forward.
(a)(ii)
For the correct answer:
Air resistance / drag / friction
The $8000\text{ N}$ opposing force is created by air particles constantly pushing against the surface of the moving aeroplane. In physics, we generally refer to this specific type of fluid friction as air resistance or drag.
(a)(iii)
For the correct answer:
Constant speed / constant velocity / zero acceleration
According to Newton’s first law of motion, if the resultant force on an object is exactly zero, its state of motion will remain unchanged. Since the aeroplane is already actively flying, a zero resultant horizontal force simply means it will continue soaring forward at a steady, constant speed without speeding up or slowing down.
(b)
For the correct answer:
$1200\text{ N cm}$
The rotational or turning effect of any given force is called its moment. This can be calculated precisely using the formula $\text{Moment} = \text{Force} \times \text{perpendicular distance from pivot}$. By multiplying the directly applied force of $60\text{ N}$ by the distance of $20\text{ cm}$, we achieve a final moment of $1200\text{ N cm}$.
Question 4
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.7.3$ — Energy resources (Parts $\mathrm{(a)(i)}$, $\mathrm{(a)(ii)}$)
• Topic $4.5.6$ — The transformer (Parts $\mathrm{(b)}$, $\mathrm{(c)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
X: turbine
Y: generator
In a standard thermal power station, the high-pressure steam generated by boiling water is directed into a massive component called a turbine (labelled X), which contains fan-like blades that spin at high speeds. This spinning turbine is mechanically linked via a shaft to a generator (labelled Y). The generator uses this rotational movement to spin magnets within coils of wire, successfully producing alternating electrical current for the grid.
(a)(ii)
For the correct answer (any two):
Chemical energy (in coal) to thermal/internal energy (in boiler)
Thermal/internal energy (of steam/water) to kinetic energy (of steam)
Kinetic energy of steam to kinetic energy of turbine/generator
Kinetic energy (of generator) to electrical energy
Power generation is fundamentally a sequence of energy conversions. Initially, the chemical energy locked inside the coal is converted into thermal energy through combustion in the boiler. This intense heat boils water, transferring thermal energy into the kinetic energy of the rapidly moving, highly pressurized steam. The fast-moving steam strikes the turbine blades, transferring its kinetic energy to the rotational kinetic energy of the turbine and generator shaft, which is finally converted into useful electrical energy.
(b)
For the correct answer:
$200000\text{ V}$
To find the output voltage of a transformer, we apply the standard transformer equation: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$. By substituting the given parameters into the formula, we get $\frac{V_s}{25000} = \frac{4800}{600}$. The ratio of turns is $8$ (since $4800 / 600 = 8$), meaning this is a step-up transformer that multiplies the voltage by a factor of eight. Therefore, $V_s = 25000 \times 8 = 200000\text{ V}$.
(c)
For the correct answer (any two):
Reduces current (in cables)
Less energy or power wasted or less heating or more efficient
Enables use of thinner cables
(so) lower cost for cable and supporting pylons
Transmit (electricity over) longer distances (without drop in p.d.)
When electrical power is transmitted across the national grid, utilizing extremely high voltages proportionally decreases the electrical current flowing through the lines ($P = IV$). Because energy loss in a wire due to heating is proportional to the square of the current ($P = I^2R$), keeping the current as low as possible drastically minimizes wasted heat. Consequently, power companies can use thinner, lighter, and cheaper cables to reliably transmit electricity over vast geographical distances without losing significant amounts of useful energy.
Question 5
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $2.3.1$ — Conduction (Parts $\mathrm{(a)(i)}$, $\mathrm{(a)(ii)}$)
• Topic $2.2.3$ — Melting, boiling and evaporation (Part $\mathrm{(b)}$)
• Topic $2.1.2$ — Particle model (Parts $\mathrm{(c)(i)}$, $\mathrm{(c)(ii)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
(shiny surfaces) are good reflectors OR poor absorbers/emitters
so less thermal energy lost by radiation
Thermal energy can be transferred through empty space in the form of infrared electromagnetic radiation. Shiny, silvered surfaces act much like a visible mirror, but for heat. They are highly effective at reflecting infrared radiation back into the liquid and are inherently poor at emitting their own radiation outward, thereby trapping the heat inside the flask.
(a)(ii)
For the correct answer:
less (heat lost by) convection
less (heat lost by) conduction
A true vacuum is a space entirely devoid of any matter, including air molecules. Because both conduction (vibrating particles bumping into each other) and convection (the physical movement of heated fluid currents) absolutely require a physical medium to transfer energy, the vacuum creates an impassable insulating barrier that prevents these two forms of heat loss entirely.
(b)
For the correct answer:
more energetic particles
particles escape (from the surface (attraction))
so average energy of particles remaining (in liquid) decreases
In any given liquid, the particles are in constant, random motion at varying speeds. During evaporation, the fastest-moving particles near the very surface possess enough kinetic energy to overcome the attractive intermolecular forces holding them in the liquid state, allowing them to escape into the air as a gas. Because the most energetic particles have departed, the average kinetic energy of the remaining liquid particles inherently decreases, which we observe as a drop in the liquid’s overall temperature.
(c)(i)
For the correct answer:
Brownian (motion)
The erratic, random, and constant zigzag movement of microscopic particles (like pollen or smoke) when suspended in a fluid is called Brownian motion. It provides critical, observable evidence for the kinetic particle theory of matter.
(c)(ii)
For the correct answer:
(fast moving liquid) molecules
bombard/collide with (small) particle
collisions produce (resultant) force (in random directions)
While the small particle is visible under the microscope, it is surrounded by millions of much smaller, invisible liquid molecules moving at incredibly high speeds. These unseen molecules constantly bombard the visible particle from all sides. In any given millisecond, there might be slightly more molecular collisions on one side than another, creating a small, instantaneous resultant force that pushes the visible particle in a random direction, causing the zigzag path.
Question 6
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $3.3$ — Electromagnetic spectrum (Parts $\mathrm{(a)(i)}$, $\mathrm{(a)(ii)}$, $\mathrm{(c)}$)
• Topic $3.1$ — General properties of waves (Parts $\mathrm{(b)}$, $\mathrm{(d)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
1st box: microwaves
2nd box: infrared
The electromagnetic spectrum follows a specific order from longest wavelength (lowest frequency) to shortest wavelength (highest frequency). Between radio waves and visible light, the spectrum consists of microwaves, followed immediately by infrared radiation.
(a)(ii)
For the correct answer:
X-rays OR gamma rays
Wavelength decreases as you move across the spectrum past visible light. Ultraviolet sits just past the violet end of visible light. Moving to even shorter wavelengths and higher energy levels, you find X-rays and finally gamma ($\gamma$) rays.
(b)
For the correct answer:
Horizontal line drawn between 2 peaks OR any 2 adjacent similar points on the wave.
A single wavelength ($\lambda$) represents one complete cycle of a wave. The easiest and clearest way to indicate this on a displacement-distance graph is to draw a straight horizontal line connecting two adjacent crests (peaks) or two adjacent troughs.
(c)
For the correct answer:
Correction 1: microwaves
Correction 2: X-rays
Mobile phones and satellite television rely on microwaves (and sometimes radio waves) because they can pass through the Earth’s atmosphere to reach satellites. For medical imaging of the skeletal system, doctors use X-rays because they pass easily through soft tissue but are absorbed by denser materials like bone, casting a shadow that creates the image.
(d)
For the correct answer:
Longitudinal (vibrations) are parallel to the direction of propagation.
Transverse (vibrations) are perpendicular/at right angles to the direction of propagation.
The defining characteristic of wave types is how the particles of the medium behave. In longitudinal waves (like sound), particles oscillate back and forth along the exact same axis the wave is travelling. In transverse waves (like water waves or light), the disturbance moves up and down or side to side, exactly $90^{\circ}$ to the direction the energy is flowing.
Question 7
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $3.2.2$ — Refraction of light (Parts $\mathrm{(a)(i)}$, $\mathrm{(a)(ii)}$)
• Topic $3.2.3$ — Thin lenses (Parts $\mathrm{(b)(i)}$, $\mathrm{(b)(ii)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
$29^{\circ}$
By absolute definition in physics, the angle of refraction is exclusively measured between the newly refracted ray and the normal line (the perpendicular dashed line) inside the denser medium. Based on the diagram provided, this specific angle is precisely $29^{\circ}$.
(a)(ii)
For the correct answer:
normal (line)
Line X represents the imaginary reference line drawn completely perpendicular ($90^{\circ}$) to the boundary surface at the exact point where the light ray strikes. This crucial reference line is universally called the normal.
(b)(i) & (ii)
For the correct answer:
Ray through centre continues in straight line.
Ray through F drawn parallel to principal axis.
Arrow drawn from principal axis to where rays cross.
To accurately construct a ray diagram for a converging lens, you track two primary rays from the tip of the object. The ray that travels directly through the optical centre passes straight through without bending. The ray that passes through the focal point (F) before hitting the lens refracts to emerge perfectly parallel to the principal axis. Where these two straight lines intersect is exactly where the tip of the real, inverted image forms. You then draw a vertical line (usually an arrow) from the principal axis down to this intersection point.
Question 8
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $4.2.1$ — Electric charge (Parts $\mathrm{(a)(i)}$, $\mathrm{(c)(i)}$, $\mathrm{(c)(ii)}$)
• Topic $4.1$ — Simple phenomena of magnetism (Parts $\mathrm{(a)(ii)}$, $\mathrm{(b)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
plastic strip AND glass lens
Electrical insulators are materials that do not possess free-moving electrons, meaning electric current cannot easily flow through them. Metals like aluminium, silver, and iron are all conductors. Non-metals like plastic and glass tightly bind their electrons, making them excellent insulators.
(a)(ii)
For the correct answer:
iron bar
While many metals conduct electricity, only a select few specific elements are naturally magnetic at room temperature (ferromagnetic). The most common examples are iron, cobalt, and nickel. Therefore, the iron bar is the only magnetic item listed.
(b)
For the correct answer:
End of magnet X labelled S (pole) AND end of magnet Y nearest magnet X labelled N (pole) AND other end is S (pole).
One of the fundamental laws of magnetism is that opposite poles attract while like poles repel. Since the diagram clearly states the magnets are pulling toward one another (attracting), the pole on magnet X facing magnet Y must be South (S). Consequently, the side of magnet Y facing X must be North (N), leaving the far end of Y as a South (S) pole.
(c)(i)
For the correct answer:
Spheres drawn closer together.
When sphere P becomes positively charged, it exerts an attractive electrostatic force on the uncharged, conductive sphere Q by pulling its free electrons closer to the side facing P (electrostatic induction). This creates a net attraction, pulling the hanging spheres visibly closer together.
(c)(ii)
For the correct answer:
Spheres drawn further apart. Both strings at an angle to vertical.
A core rule of static electricity is that like charges repel. Because both sphere P and sphere Q have been given the same positive charge, they will actively push each other away. You must draw both cotton threads angled outward from the vertical, with the spheres hanging noticeably farther apart than in their original resting state.
Question 9
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $4.3.1$ — Circuit diagrams and circuit components (Part $\mathrm{(a)}$)
• Topic $4.2.4$ — Resistance (Parts $\mathrm{(b)(i)}$, $\mathrm{(b)(ii)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
Correct symbol for battery.
Correct symbol for switch.
Correct symbol for lamp.
All 3 components connected in series.
A standard circuit diagram requires precise, globally recognized symbols. You must draw a battery (a series of long and short parallel lines), an open or closed switch, and a lamp (a circle containing an ‘X’). Because the torch works as a single continuous loop where the current has only one path to flow, all three of these components must be drawn connected consecutively in a simple series circuit.
(b)(i)
For the correct answer:
$0.26\text{ A}$
In a standard series circuit, the electrical current is conserved and remains completely identical at absolutely every point along the path. Because the brass connecting strip is in series with the lamp, the current flowing through it is exactly the same as the current flowing through the lamp: $0.26\text{ A}$.
(b)(ii)
For the correct answer:
$5.4\ \Omega$
Electrical resistance is defined by Ohm’s law, which states $R = \frac{V}{I}$. By substituting the given potential difference ($1.4\text{ V}$) and the current ($0.26\text{ A}$), we calculate $R = \frac{1.4}{0.26}$, which yields approximately $5.38\ \Omega$. This reliably rounds to two significant figures as $5.4\ \Omega$.
Question 10
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $4.5.3$ — Magnetic effect of a current (Part $\mathrm{(a)}$)
• Topic $4.5.4$ — Force on a current-carrying conductor (Parts $\mathrm{(b)(i)}$, $\mathrm{(b)(ii)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
Circles drawn.
Concentric (by eye) with wire.
Arrow drawn clockwise on/near field (line).
Whenever an electric current flows through a straight conductor, it generates a magnetic field in the shape of concentric circles radiating outward. To determine the exact direction of these circular field lines, you use the right-hand grip rule: pointing your thumb in the direction of the current (downwards into the card) causes your fingers to curl in a clockwise direction. Thus, you must draw circles around the wire with clear clockwise arrows.
(b)(i)
For the correct answer (any two):
Increase current (in wire)
Increase strength of magnets or magnetic field
The magnetic force acting on a current-carrying wire placed in an external magnetic field (often called the motor effect) is directly proportional to both the current and the magnetic field strength. Therefore, ramping up the electrical current or using significantly stronger permanent magnets will correspondingly increase the physical upward push experienced by the wire.
(b)(ii)
For the correct answer:
Reverse the (direction of the) current (in the wire).
Reverse the magnetic field (or move poles closer together, though reversing is the standard method).
Fleming’s Left-Hand Rule dictates the directional relationship between current, magnetic field, and the resulting force. Because these three vectors are perfectly perpendicular, flipping the direction of exactly one of the inputs (either making the current flow from Y to X, or physically swapping the North and South magnet poles) will precisely invert the direction of the resulting force, making it push downwards instead of upwards.
Question 11
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $5.1.2$ — The nucleus (Part $\mathrm{(a)}$)
• Topic $5.2.4$ — Half-life (Part $\mathrm{(b)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
$^{241}_{95}\text{Am}$
In standard scientific nuclide notation ($^{A}_{Z}\text{X}$), the bottom number $Z$ represents the proton number (atomic number), which defines the element. The top number $A$ represents the total nucleon number (mass number), which is the sum of protons and neutrons. For this americium isotope, $Z = 95$, and $A = 95 + 146 = 241$. Placing these correctly around the chemical symbol Am gives $^{241}_{95}\text{Am}$.
(b)
For the correct answer:
$430\text{ years}$
A radioactive substance’s half-life is strictly defined as the exact time required for its measured activity (or count rate) to decay to half of its initial value. Looking at the y-axis of the provided graph, the decay curve begins at an initial count rate of $16000\text{ counts/min}$. Half of this value is $8000\text{ counts/min}$. If you trace a horizontal line from $8000$ to intercept the decay curve, and then trace vertically downwards to the x-axis, it aligns perfectly at $430\text{ years}$.