Question 1
Calculate the density of the steel.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.1 — Physical quantities and measurement techniques (Parts (a)(i), (a)(ii))
• Topic 1.4 — Density (Part (b))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
$4.3\text{ cm}$
To find the distance AB, you need to read the positions of points A and B on the ruler provided in the image. Identify the reading at the starting point A, which aligns exactly with the $1.5\text{ cm}$ mark on the ruler’s scale. Next, identify the reading at the ending point B, which aligns perfectly with the $5.8\text{ cm}$ mark. The total distance AB is the difference between these two positional readings. Subtracting the initial reading from the final reading gives the calculation: $5.8\text{ cm} – 1.5\text{ cm} = 4.3\text{ cm}$.
(a)(ii)
For the correct answer:
$0.54\text{ cm}$
The distance AB calculated in the previous step represents the total combined length of all the steel balls lined up in a straight row. By carefully counting the identical steel balls positioned between point A and point B in the diagram, you will find there are exactly $8$ balls. Since all the balls are identical, each has the exact same diameter. To find the diameter of a single steel ball, divide the total distance by the total number of balls. This gives $4.3\text{ cm} \div 8 = 0.5375\text{ cm}$. Rounding this value to two significant figures, as is standard practice for this level of measurement precision, yields a final diameter of $0.54\text{ cm}$.
(b)
For the correct answer:
$7.8\text{ g/cm}^3$
Density is an intrinsic physical property of a material, defined as its total mass per unit volume. The standard formula used to calculate density is $\rho = \frac{m}{V}$, where ‘$m$’ represents the mass and ‘$V$’ represents the volume. The question provides the total mass of the given steel balls as $54\text{ g}$ and their total combined volume as $6.9\text{ cm}^3$. Substituting these specific values into our density formula gives $\rho = \frac{54\text{ g}}{6.9\text{ cm}^3}$. Performing this division results in approximately $7.82608\text{ g/cm}^3$. Rounding this to two significant figures (to match the precision of the given data), we determine the density of the steel is $7.8\text{ g/cm}^3$.
Question 2
(i) constant speed
(ii) constant deceleration
(iii) constant non-zero acceleration
(i) Calculate the size of the resultant force on the cyclist.
(ii) State the effect, if any, of the resultant force on the motion of the cyclist.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.2 — Motion (Parts (a)(i), (a)(ii), (a)(iii), (b))
• Topic 1.5.1 — Effects of Forces (Parts (c)(i), (c)(ii))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
ST or WX
In a speed-time graph, a horizontal line with a gradient of zero indicates that the speed is remaining steady over time. Both sections ST and WX show the graph as a flat, horizontal line, meaning the cyclist is moving at a constant speed during these specific time intervals without speeding up or slowing down.
(a)(ii)
For the correct answer:
XY
Deceleration is represented by a negative gradient (or downward slope) on a speed-time graph. Section XY shows a straight line sloping downwards, indicating a uniform and steady decrease in speed over time, which corresponds directly to constant deceleration.
(a)(iii)
For the correct answer:
TW or XY
Acceleration is determined by the gradient of a speed-time graph. Section TW has a constant positive gradient (representing constant acceleration), while section XY has a constant negative gradient (representing constant deceleration). Both of these sections represent a constant, non-zero rate of change in velocity.
(b)
For the correct answer:
100 m
The total distance travelled by an object can be calculated by finding the area under the speed-time graph for that specific section. Section ST forms a rectangular shape on the graph. By multiplying the constant speed (the vertical height) by the time duration (the horizontal width) for section ST, you obtain the total distance covered, which equals 100 m.
(c)(i)
For the correct answer:
60 N
The resultant force is determined by calculating the difference between the opposing horizontal forces acting on the object. The forward driving force is given as 100 N, and the opposing backward force is 40 N. Subtracting the backward force from the forward force (100 N – 40 N) leaves a net resultant force of 60 N acting forwards.
(c)(ii)
For the correct answer:
The cyclist accelerates or increases speed.
According to Newton’s laws of motion, an unbalanced, non-zero resultant force causes an object to change its velocity. Since the 60 N resultant force acts in the exact same direction as the cyclist’s current motion, it provides a forward push that causes the cyclist to accelerate, resulting in a continuous increase in speed.
Question 3
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.7$ — Energy, work and power (Parts $\mathrm{(a)}$, $\mathrm{(b)}$)
• Topic $1.8$ — Pressure (Part $\mathrm{(c)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
Electrical energy → Chemical energy (100 J)
Chemical energy → Light energy (70 J)
Chemical energy → Thermal energy (30 J)
In a battery-powered torch, the primary energy store is the chemical energy within the battery. When switched on, this chemical energy is transferred into electrical energy by the circuit. The electrical energy is then transferred by the bulb into useful light energy to illuminate the surroundings, and wasted thermal (heat) energy due to the resistance of the filament. According to the principle of conservation of energy, the total input energy equals the sum of the output energies.
(b)
For the correct answer:
6.8 J
Work done is calculated when a force moves an object over a distance in the direction of the force. The formula is $W = Fd$, where $W$ is work done, $F$ is the force applied (the 8.5 N weight of the torch), and $d$ is the vertical distance moved. Substituting the given values: $W = 8.5 \times 0.80$. This gives a total work done of 6.8 J, which is also equal to the gravitational potential energy gained by the torch.
(c)
For the correct answer:
0.19 N/cm2
Pressure is defined as the force exerted per unit area. It can be calculated using the equation $p = \frac{F}{A}$, where $F$ is the force (the weight of the torch) and $A$ is the contact area of the base. Substituting the given values into the formula gives $p = \frac{8.5}{44}$. This calculates to approximately 0.19 N/cm2, indicating how the weight is distributed over the shelf’s surface.
Question 4
(i) Describe the arrangement, separation, and motion of the particles in the solid metal.
(ii) The student cools the block of metal in a freezer. State the effect, if any, of cooling on the kinetic energy of the particles in the block of metal.
(i) State the name of the temperature at which particles have the least kinetic energy.
(ii) State the value of temperature at which particles have the least kinetic energy. Include the unit.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 2.1.2 — Particle model (Parts (a)(i), (a)(ii), (b)(i), (b)(ii))
• Topic 2.3.3 — Radiation (Part (c))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
• Particles are fixed in position.
• Particles are arranged in a regular pattern.
• Particles vibrate about fixed positions.
• Particles are close together.
In a solid, particles are arranged in a regular, closely packed lattice structure with very little separation between them. Because they are tightly bound together by strong intermolecular forces, they cannot move freely from place to place but only vibrate around their fixed positions.
(a)(ii)
For the correct answer:
The kinetic energy of the particles decreases.
Temperature is a direct measure of the average kinetic energy of the particles in a substance. As the metal block cools down in the freezer, the particles lose thermal energy, meaning they vibrate less vigorously and their kinetic energy decreases.
(b)(i)
For the correct answer:
Absolute zero
The lowest possible temperature in the universe is known as absolute zero. At this theoretical point, all thermal energy is removed, and particles reach their minimum possible kinetic energy state with negligible motion.
(b)(ii)
For the correct answer:
-273 °C or 0 K
Absolute zero corresponds to exactly 0 K on the Kelvin thermodynamic temperature scale. When converted to the Celsius scale, this is equivalent to approximately -273 °C.
(c)
For the correct answer:
1. Black or dark in colour.
2. Dull or rough (matte) in texture.
The rate at which a surface emits thermal radiation depends heavily on its physical characteristics. Matte (dull) and black or dark-coloured surfaces are the most effective at emitting infrared radiation, whereas shiny or light-coloured surfaces tend to reflect radiation rather than emit or absorb it.
Question 5
The speed of the sound is 340 m/s.
Suggest how the quieter sound waves reach the observer.
The speed of the sound is 340 m/s.
Calculate the frequency of the sound.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 3.4 — Sound (Parts (a)(i), (a)(ii), (b))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
610 m
The distance travelled by the sound wave can be calculated using the standard formula relating speed, distance, and time: distance = speed × time. By substituting the given values, the distance is calculated as 340 m/s × 1.8 s = 612 m. To maintain consistency with the two significant figures provided in the time measurement (1.8 s), the final distance is rounded appropriately to 610 m.
(a)(ii)
For the correct answer:
The quieter sound waves reach the observer as an echo. The sound waves reflect off the rocks or the bottom of the quarry (YZ or Z) and travel back to the observer.
Sound waves possess the ability to reflect off hard, rigid surfaces, such as the rock walls or the floor of the quarry. When the initial sound waves from the explosion hit these surfaces, they bounce back towards the observer, creating an echo. This reflected sound is perceived as quieter because the waves have travelled a greater total distance, causing the sound energy to spread out and dissipate over the longer journey.
(b)
For the correct answer:
1200 Hz
The relationship between the speed, frequency, and wavelength of a wave is defined by the wave equation $v = f \lambda$. To find the frequency ($f$), the formula can be rearranged to $f = \frac{v}{\lambda}$. Substituting the known values into the equation yields $f = \frac{340}{0.28}$, which calculates to approximately 1214 Hz. Rounding this result to two significant figures, consistent with the given wavelength measurement, provides a final frequency of 1200 Hz.
Question 6
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $3.1$ — General properties of waves (Part $\mathrm{(a)}$)
• Topic $3.2$ — Light (Parts $\mathrm{(a)}$, $\mathrm{(b)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
Refraction
The process shown is refraction. This occurs when light waves travel from one medium (air) into another medium with a different optical density (glass). As the waves enter the glass block, their transmission speed changes. This change in speed causes the path of the light to bend, which is clearly indicated by the change in the direction and spacing of the wavefronts in the diagram.
(a)(ii)
For the correct answer:
1. The speed of the light waves decreases.
2. The wavelength of the light waves decreases.
When light waves pass from a less dense medium like air into a more optically dense medium like glass, they slow down, meaning their speed decreases. Because the frequency of the light wave must remain constant (as it is determined purely by the source), the wave equation $v = f\lambda$ dictates that the wavelength must also decrease proportionally to the drop in speed.
(b)
For the correct answer:
The ray of red light undergoes total internal reflection as it travels through the glass fibre. This occurs because the light is travelling from a more dense medium (glass) into a less dense medium (air) at an angle greater than the critical angle. The light reflects off the inner surface of the glass fibre and continues to travel along the fibre.
The ray of red light propagates through the solid glass fibre by undergoing a phenomenon called total internal reflection. Because the light is travelling from a medium with a higher refractive index (glass) towards a boundary with a lower refractive index (air), and it strikes the boundary at an angle of incidence that is greater than the critical angle, no light refracts outwards. Instead, the light is reflected entirely back into the core, allowing it to travel continuously down the length of the fibre without escaping.
Question 7
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $4.1$ — Simple phenomena of magnetism (Parts $\mathrm{(a)(i)}$, $\mathrm{(a)(ii)}$)
• Topic $4.5.1$ — Electromagnetic induction (Parts $\mathrm{(b)(i)}$, $\mathrm{(b)(ii)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
Induced magnetism occurs when a magnetic material becomes magnetised when placed in a magnetic field.
Magnetic materials like iron or steel do not initially act as magnets, but when brought near a permanent magnet, the external field aligns their internal domains. This cause the material to temporarily acquire its own magnetic poles, allowing it to attract other magnetic objects just like a permanent magnet does.
(a)(ii)
For the correct answer:
Iron nails lose magnetism quickly while steel nails retain it.
Soft magnetic materials like iron are easily magnetised but lose their magnetism almost immediately once the external field is removed. In contrast, hard magnetic materials like steel require more effort to magnetise but retain their magnetic strength for a long time, effectively becoming permanent magnets themselves.
(b)(i)
For the correct answer:
Zero (0) reading.
A voltmeter only shows a reading if there is a change in the magnetic flux linkage or if the conductor is actively cutting through magnetic field lines. Since the wire XY is stationary relative to the magnet, no field lines are being cut and no electromotive force (e.m.f.) is induced, resulting in no potential difference.
(b)(ii)
For the correct answer:
The wire cuts through magnetic field lines, inducing an e.m.f.
When the wire moves perpendicularly to the magnetic field, it interrupts the magnetic flux lines existing between the North and South poles. According to Faraday’s Law, this relative motion induces an electromotive force (e.m.f.) across the ends of the wire, which the voltmeter detects as a non-zero voltage reading.
Question 8
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $4.2$ — Electrical quantities (Parts $\mathrm{(a)}$, $\mathrm{(b)}$, $\mathrm{(c)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
Ammeter (A) in series; Voltmeter (V) in parallel.
In a circuit, an ammeter must be connected in series with the component to measure the current flowing through it, while a voltmeter must be connected in parallel to measure the potential difference across it. Therefore, the circle in the main loop should be labeled ‘A’ and the circle branching across the heater should be labeled ‘V’.
(b)
For the correct answer:
$5.7\text{ }\Omega$
Resistance is calculated using Ohm’s Law, defined by the formula $R = \frac{V}{I}$. By substituting the given values of potential difference ($V = 8.0\text{ V}$) and current ($I = 1.4\text{ A}$), we get $R = \frac{8.0}{1.4} \approx 5.714\text{ }\Omega$. Rounding to two significant figures, as provided in the question’s data, yields a final resistance of $5.7\text{ }\Omega$.
(c)
For the correct answer:
$336\text{ J}$ (or $340\text{ J}$)
Electrical energy transferred is determined by the formula $E = IVt$, where $I$ is current, $V$ is potential difference, and $t$ is time in seconds. Substituting the values $1.4\text{ A}$, $8.0\text{ V}$, and $30\text{ s}$ into the equation gives $E = 1.4 \times 8.0 \times 30 = 336\text{ J}$. This can be rounded to $340\text{ J}$ to maintain consistency with two significant figures.
Question 9
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $4.4$ — Electrical safety (Part $\mathrm{(a)}$)
• Topic $4.5$ — Electromagnetic effects (Part $\mathrm{(b)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
Live wire; because the live wire is at a high potential ($240\text{ V}$), placing the switch here ensures that when the circuit is open, no part of the appliance remains at a high voltage, preventing electric shocks.
(b)(i)
For the correct answer:
Step-down transformer
(b)(ii)
For the correct answer:
A step-down transformer consists of a primary coil and a secondary coil wound around a soft iron core. To reduce voltage, the primary coil has significantly more turns than the secondary coil. The soft iron core is used because it is easily magnetised and demagnetised, which efficiently links the changing magnetic flux from the primary to the secondary coil to induce the required output voltage.
Question 10
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 5.1.2 — The nucleus (Parts (a)(i), (a)(ii))
• Topic 5.2.2 — The three types of nuclear emission (Part (b))
• Topic 5.2.4 — Half-life (Part (c))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
53
In nuclide notation Z A X, the bottom number (Z) represents the proton number (atomic number). For Iodine-131, this value is 53, which directly indicates the number of protons present in the nucleus.
(a)(ii)
For the correct answer:
78
The top number (A) is the nucleon number, representing the sum of protons and neutrons. To find the neutron count, subtract the proton number (Z) from the nucleon number (A): 131−53=78 neutrons.
(b)
For the correct answer:
Beta-particle: high-energy electron; Gamma ray: high-energy electromagnetic wave
A beta (β)-particle is a fast-moving electron emitted from the nucleus when a neutron decays into a proton. A gamma (γ) ray is not a particle but a high-frequency electromagnetic wave, often emitted alongside α or β radiation to release excess energy from the nucleus.
(c)
For the correct answer:
0.2 mg
First, determine the number of half-lives by dividing the total time by the half-life duration: 24.0/8.0=3 half-lives. The mass halves with each period: after 8 days it is 0.8 mg, after 16 days it is 0.4 mg, and finally, after 24 days, the remaining mass is 0.2 mg.
Question 11
Fig. 11.1 shows the Sun and the four innermost planets, A, B, C, and D, of the Solar System.
(a) In Table 11.1, write the names of the innermost planets. One is done for you.
| Planet | Name of Planet |
|---|---|
| A | …… |
| B | Venus |
| C | …… |
| D | …… |
(b) Describe how the four innermost planets of the Solar System were formed.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $6.1.1$ — The Earth (Part $\mathrm{(a)}$
• Topic $6.1.2$ — The Solar System (Part $\mathrm{(b)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
A: Mercury, C: Earth, D: Mars
The Solar System is organized by the distance of planets from the Sun. Mercury (A) is the closest, followed by Venus (B), then Earth (C), and finally Mars (D) as the fourth innermost planet. These four are known as the terrestrial or rocky planets due to their solid compositions compared to the gas giants further out.
(b)
For the correct answer:
The planets formed from the accretion of dust and gas in a rotating nebula.
The innermost planets originated from a giant rotating cloud of dust and gas called the solar nebula. As the nebula collapsed under gravity, it flattened into a disk with the young Sun at the center. In the hot inner regions, solid grains of rock and metal collided and stuck together through accretion to form planetesimals, which eventually merged into the four rocky planets.