Question 1
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 1.2 — Motion (Parts (a)(i), (a)(ii), (b), (c), (d))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
18 m/s
Identify the highest point on the speed-time graph. In Fig. 1.1, the maximum speed occurs between 10 s and 30 s, where the line is at its peak horizontal level. Reading across to the vertical speed axis, the value is exactly 18 m/s.
(a)(ii)
For the correct answer:
12 m/s
To find the speed at a specific time, locate 65 s on the horizontal axis and draw a vertical dashed line up to the curve. From that intersection point, draw a horizontal dashed line across to the vertical axis. The value on the speed axis corresponds to 12 m/s.
(b)
For the correct answer:
The speed decreases/decelerates until 75 s and then remains at 0 m/s (stationary).
Between 60 s and 75 s, the negative gradient indicates constant deceleration as the speed drops from 18 m/s to zero. From 75 s to 80 s, the line lies on the horizontal axis (0 m/s), showing that the bus has come to a complete stop and is stationary.
(c)
For the correct answer:
90 m
The distance travelled is represented by the area under the speed-time graph. For the first 10 s, the shape is a triangle with a base of 10 s and a height of 18 m/s. Using the formula Area= 2 1 ×base×height, the distance is 2 1 ×10 s×18 m/s=90 m.
(d)
For the correct answer:
A straight line drawn from (40, 14.0) to (45, 6.0).
Starting at 40 s, the initial speed is 14.0 m/s. Since the speed reduces over 5.0 s, the end point is at 45 s (40+5). At this time, the speed is 6.0 m/s. A straight diagonal line connecting these two points represents the uniform deceleration during braking.
Question 2
(i) State the energy store of the barrel of fruit that increases as the barrel rises.
(ii) The weight of the barrel of fruit is $140\text{ N}$. Show that the work done on the barrel of fruit in lifting it from the ground to the platform is approximately $450\text{ J}$.
(i) The output power of the machine is $75\text{ W}$. The work done in lifting a barrel of fruit onto the platform is $450\text{ J}$. Calculate the time taken for the machine to lift a barrel of fruit onto the platform.
(ii) The machine uses an electric motor. The farmer installs some wind turbines to supply electrical power for the farm. Suggest one environmental reason for using wind turbines rather than using a diesel (fossil fuel) generator.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.7.1$ — Energy (Part $\mathrm{(a)(i)}$)
• Topic $1.7.2$ — Work (Part $\mathrm{(a)(ii)}$)
• Topic $1.7.4$ — Power (Part $\mathrm{(b)(i)}$)
• Topic $1.7.3$ — Energy resources (Part $\mathrm{(b)(ii)}$)
▶️ Answer/Explanation
(a)(i)
For the correct answer:
gravitational potential energy
When an object is lifted vertically, work is done against the force of gravity to increase its height. According to the syllabus, this energy is stored as gravitational potential energy. Since the mass and gravity are constant, the increase in vertical distance directly results in a higher energy store within the barrel.
(a)(ii)
For the correct answer:
$W = F \times d = 140 \times 3.2 = 448\text{ J}$ (which is $\approx 450\text{ J}$)
Mechanical work done is defined by the equation $W = Fd$, representing the product of the force applied and the distance moved in the direction of that force. By substituting the barrel’s weight ($140\text{ N}$) as the force and the platform height ($3.2\text{ m}$) as the distance, the calculation yields $448\text{ J}$, confirming the approximation.
(b)(i)
For the correct answer:
$6.0\text{ s}$
Power is the rate at which work is done, expressed by the formula $P = \frac{W}{t}$. To find the time taken, the equation is rearranged to $t = \frac{W}{P}$. Dividing the work done ($450\text{ J}$) by the machine’s power output ($75\text{ W}$) results in a duration of exactly $6$ seconds to complete the lift.
(b)(ii)
For the correct answer:
Wind is a renewable source OR does not produce greenhouse gases/$\text{CO}_2$ OR does not contribute to global warming
Wind turbines utilize a renewable energy source that does not deplete over time, unlike fossil fuels used in diesel generators. Environmentally, turbines are superior because they generate electricity without burning fuel, thereby avoiding the emission of carbon dioxide and other harmful pollutants that drive climate change and air pollution.
Question 3
The mass of a glass bottle is $0.18\text{ kg}$.
(a) Calculate the weight of the bottle.
(b) The bottle contains $2.7\text{ kg}$ of cooking oil. The density of the cooking oil is $0.92\text{ g/cm}^3$. Calculate the volume of the cooking oil.
(c) A cookery student pours some cooking oil into a glass bowl containing water, as shown in Fig. 3.1.
The student accidentally drops a plastic spoon and a metal spoon into the bowl. The densities of the spoons and liquids are shown in Table 3.1.
On Fig. 3.1, label a suggested position for each spoon after each has fallen into the bowl.
Use the letter P to label the position of the plastic spoon and the letter M to label the position of the metal spoon.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $1.3$ — Mass and weight (Part $\mathrm{(a)}$)
• Topic $1.4$ — Density (Parts $\mathrm{(b)}$, $\mathrm{(c)}$)
▶️ Answer/Explanation
(a)
For the correct answer:
$1.8\text{ N}$
Weight is the gravitational force acting on an object and is calculated using the formula $W = m \times g$. Given the mass $m = 0.18\text{ kg}$ and using the standard gravitational field strength $g = 9.8\text{ N/kg}$ (or approximately $10\text{ N/kg}$), the calculation is $0.18\text{ kg} \times 9.8\text{ m/s}^2 = 1.764\text{ N}$. When rounded to two significant figures, as is standard for IGCSE physics based on the provided values, the result is $1.8\text{ N}$.
(b)
For the correct answer:
$2900\text{ cm}^3$
Volume is derived from the density formula $\rho = \frac{m}{V}$, rearranged to $V = \frac{m}{\rho}$. To ensure consistent units, the mass of the oil must be converted from kilograms to grams ($2.7\text{ kg} = 2700\text{ g}$). Dividing this mass by the density gives $2700\text{ g} \div 0.92\text{ g/cm}^3 \approx 2934.78\text{ cm}^3$, which rounds to $2900\text{ cm}^3$ to maintain appropriate precision.
(c)
For the correct answer:
P on top liquid surface, M on bottom of bowl
Floating and sinking are determined by comparing the density of an object to the density of the surrounding fluid. The plastic spoon ($0.76\text{ g/cm}^3$) is less dense than both the oil ($0.92\text{ g/cm}^3$) and water ($1.0\text{ g/cm}^3$), meaning it will float at the very top. Conversely, the metal spoon ($8.7\text{ g/cm}^3$) is significantly denser than both liquids, causing it to sink through both layers to the base of the container.
Question 4
(i) a wooden handle
(ii) shiny sides
(iii) a copper base.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 2.3.1 — Conduction (Part (a)(i), (a)(iii))
• Topic 2.3.2 — Convection (Part (b))
• Topic 2.3.3 — Radiation (Part (a)(ii))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
Wood is a poor conductor (insulator) of heat.
Thermal conduction involves the transfer of energy through a material. Wood is a thermal insulator, meaning it does not allow heat to flow easily. This prevents the handle from reaching high temperatures, ensuring the user can safely lift the pan without suffering burns from the thermal energy of the hotplate.
(a)(ii)
For the correct answer:
Shiny surfaces are poor emitters of infrared radiation.
Thermal energy is lost to the environment through radiation. Shiny or light-colored surfaces are inefficient at emitting infrared radiation compared to dull, black surfaces. By having shiny sides, the pan minimizes the rate of energy transfer to the surrounding air, keeping the internal contents hot for a longer period.
(a)(iii)
For the correct answer:
Copper is a very good thermal conductor.
Copper has a high rate of thermal conduction, often due to the movement of free electrons within its metallic lattice. A copper base ensures that thermal energy from the hotplate is transferred rapidly and uniformly to the water inside, increasing the efficiency of the cooking process through direct contact.
(b)
For the correct answer:
Air expands, becomes less dense, and rises, creating a convection current.
As air near the heater is warmed, the particles gain kinetic energy and move further apart, causing the air to expand and its density to decrease. This less dense, warm air rises, while cooler, denser air sinks to take its place. This continuous cycle forms a convection current that circulates thermal energy throughout the entire volume of the room.
Question 5
(ii) State one harmful effect of excessive exposure to infrared radiation.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 3.3 — Electromagnetic spectrum (Parts (a)(i), (a)(ii), (b)(i), (b)(ii))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
Left box: Ultraviolet; Right box: Microwaves
The electromagnetic spectrum is organized by wavelength; ultraviolet sits between X-rays and visible light, while microwaves are located between infrared and radio waves. As wavelength increases from left to right, the frequency decreases accordingly.
(a)(ii)
For the correct answer:
They are all transverse waves AND they all travel at the same speed in a vacuum (3.0×10 8 m/s).
All electromagnetic waves consist of oscillating electric and magnetic fields that are perpendicular to the direction of energy transfer, classifying them as transverse. In a vacuum, every region of the spectrum propagates at the universal speed of light, regardless of its frequency or wavelength.
(b)(i)
For the correct answer:
Thermal imaging OR remote controllers (for televisions) OR electric grills.
Infrared radiation is primarily utilized for its heating effects and communication capabilities. It is commonly used in household remote controls to send signals via pulses of light and in thermal imaging cameras to detect heat signatures from objects or people.
(b)(ii)
For the correct answer:
Skin burns.
Excessive exposure to infrared radiation leads to the absorption of energy by the skin’s surface, which significantly increases the temperature of the tissue. This rapid heating can damage cells and result in painful thermal skin burns.
Question 6
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 3.2.2 — Refraction of light (Part $\mathrm{(a)}$)
• Topic 3.2.3 — Thin lenses (Part $\mathrm{(b)}$)
• Topic 3.2.4 — Dispersion of light (Part $\mathrm{(c)}$)
▶️ Answer/Explanation
(a)(i)
The normal is an imaginary construction line drawn perpendicular (at $90^{\circ}$) to the boundary surface at the exact point where the incident ray strikes the glass block. It serves as the reference for measuring the angles of incidence, reflection, and refraction. On Fig 6.1, this should be a dashed vertical line passing through the point of entry.
(a)(ii)
The ray striking the block from the air is the incident ray. The ray that bounces back into the air from the first surface is the reflected ray. The ray that enters the glass and changes direction due to a change in optical density (slowing down in the glass) is the refracted ray.
(b)
In Fig. 6.2, the converging (convex) lens causes parallel rays to refract inward and meet at a single point called the principal focus (focal point). In Fig. 6.3, the diverging (concave) lens causes parallel rays to refract outward (spread apart) so they appear to originate from a virtual principal focus located behind the lens on the same side as the source.
(c)
The seven colours of the visible spectrum in order of increasing frequency (lowest to highest) are: Red, Orange, Yellow, Green, Blue, Indigo, and Violet. While Red has the longest wavelength and lowest frequency, Violet has the shortest wavelength and highest frequency; this sequence can be easily remembered using the acronym ROYGBIV.
Question 7
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic $4.3.2$ — Series and parallel circuits (Parts $\mathrm{(a)}$, $\mathrm{(e)}$)
• Topic $4.2.2$ — Electric current (Part $\mathrm{(b)}$)
• Topic $4.2.3$ — Electromotive force and potential difference (Part $\mathrm{(c)}$)
• Topic $4.2.5$ — Electrical energy and electrical power (Part $\mathrm{(d)}$)
▶️ Answer/Explanation
(a)
For the correct answer: $100\ \Omega$
In a series circuit, the total resistance is the sum of individual resistances. Since the resistors are connected end-to-end in a single loop, the combined resistance is calculated as $R_{total} = R_1 + R_2 = 50\ \Omega + 50\ \Omega = 100\ \Omega$.
(b)
For the correct answer: $0.24\ \text{A}$ (which is approx. $0.25\ \text{A}$)
Using Ohm’s Law, the current $I$ is equal to the voltage $V$ divided by the total resistance $R$. With a battery voltage of $24\ \text{V}$ and a total resistance of $100\ \Omega$, the calculation is $I = 24\ \text{V} / 100\ \Omega = 0.24\ \text{A}$, which rounds to approximately $0.25\ \text{A}$.
(c)
For the correct answer: $12\ \text{V}$
The potential difference across a single resistor in series is found using $V = I \times R$. Substituting the circuit current of $0.24\ \text{A}$ and the resistance of $50\ \Omega$ gives $V_1 = 0.24\ \text{A} \times 50\ \Omega = 12\ \text{V}$, representing half the total supply voltage.
(d)
For the correct answer: $2.88\ \text{W}$
Power is the rate of energy transfer, calculated using the formula $P = I^2R$ or $P = V \times I$. Using the values for $R_1$, the power is $P = (0.24\ \text{A})^2 \times 50\ \Omega = 0.0576 \times 50 = 2.88\ \text{W}$, indicating the energy dissipated as heat per second.
(e)
For the correct answer: Resistors connected in parallel.
To minimize combined resistance, the components must be connected in parallel so the current has multiple paths. The circuit diagram should show $R_1$ and $R_2$ placed in separate branches, both connected directly across the terminals of the $24\ \text{V}$ battery.
Question 8
Calculate the number of turns on the secondary coil of the step-up transformer. Use the information given in Fig. 8.1.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 4.5 — Electromagnetic effects (Parts (a), (b))
▶️ Answer/Explanation
(a)
For the correct answer:
7200
To find the secondary turns, we apply the transformer ratio formula: V s V p = N s N p . From the diagram, the primary voltage V p is 25 kV and the secondary voltage V s is 360 kV. Substituting the known values (N p =500) into the rearranged formula N s =N p ×( V p V s ) gives 500×( 25 360 ), resulting in 7200 turns on the secondary coil.
(b)
For the correct answer:
Reduced power loss AND improved efficiency
Transmitting electricity at high voltages significantly reduces the current flowing through the cables for a given power level (P=IV). Since power loss in transmission lines is proportional to the square of the current (P=I 2 R), a lower current minimizes heat dissipation in the wires. This makes the overall distribution system much more efficient and cost-effective over long distances.
Question 9
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 4.2.2 — Electric current (Part (a))
• Topic 4.5.3 — Magnetic effect of a current (Parts (b)(i), (b)(ii))
▶️ Answer/Explanation
(a)
For the correct answer:
current that periodically reverses its direction
Alternating current (a.c.) is characterized by a flow of electric charge that changes direction at regular intervals, unlike direct current (d.c.) which flows continuously in one direction. In a circuit, this means the polarity of the voltage source switches back and forth, causing the electrons to oscillate rather than flow in a single path.
(b)(i)
For the correct answer:
the alternating current creates a changing magnetic field around the coil which interacts with the field of the permanent magnet to produce a varying force
When an alternating current passes through the coil, it generates a magnetic field that constantly reverses its poles. This dynamic field interacts with the stationary magnetic field of the permanent magnet, creating alternating forces of attraction and repulsion (the motor effect). These rapidly changing forces cause the coil, and the attached paper cone, to move back and forth, creating the vibrations necessary to produce sound waves.
(b)(ii)
For the correct answer:
any value between 20 Hz and 20 kHz (e.g., 100 Hz or 1 kHz)
Since the teacher hears a sound, the frequency of the alternating current must correspond to the frequency of the resulting vibrations within the human audible range. Humans can typically hear frequencies from 20 Hz to 20,000 Hz. A common example for such a demonstration would be 440 Hz or 1000 Hz (1 kHz), which produces a clear, audible pitch.
Question 10
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 5.1.1 — The atom (Part (a)(iii))
• Topic 5.1.2 — The nucleus (Parts (a)(i), (a)(ii))
• Topic 5.2.4 — Half-life (Part (b))
▶️ Answer/Explanation
(a)(i)
For the correct answer:
3
The proton number, also known as the atomic number, is defined by the total number of protons located within the nucleus of an atom. By examining Fig. 10.1, we can count exactly three protons (often represented as the positively charged particles in the center), which identifies the element as lithium.
(a)(ii)
For the correct answer:
7
The nucleon number, or mass number, is the combined sum of protons and neutrons found in the atomic nucleus. According to the diagram, the lithium atom contains 3 protons and 4 neutrons; adding these together (3+4) gives a total nucleon number of 7.
(a)(iii)
For the correct answer:
It loses an electron.
An atom becomes a positive ion when it loses one or more of its negatively charged electrons from its outer shells. Since protons are positively charged and electrons are negatively charged, removing an electron leaves the lithium atom with a net positive charge, specifically forming a Li + ion.
(b)
For the correct answer:
24 days
To find the total time, determine how many half-lives occur as 80 mg decays to 10 mg: 80→40 (1 half-life), 40→20 (2 half-lives), and 20→10 (3 half-lives). Since each half-life is 8 days, the calculation is 3×8=24 days.
Question 11
(ii) State the time taken by the Earth to complete one orbit of the Sun. Include the unit.
(iii) State the time taken by the Earth to rotate once on its axis. Include the unit.
(iv) The Sun consists mainly of two gases. State the names of the two gases.
(ii) The speed of visible light is ( 3.0 \times 10^8 , \text{m/s} ). Calculate the time taken for visible light to travel from the Sun to the Earth when the Earth is in the position shown in Fig. 11.1.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• Topic 6.1.1 & 6.1.2 — The Earth and the Solar System (Parts (a)(i), (a)(ii), (a)(iii), (a)(iv))
• Topic 6.2.2 — Stars (Parts (b)(i), (b)(ii))
▶️ Answer/Explanation
(a)(i) Gravitational force. Gravity provides the centripetal force required to maintain the Earth’s elliptical orbit around the Sun, preventing it from moving off in a straight line.
(a)(ii) 365 days (or 365.25 days). This duration represents one tropical year, which is the time required for the Earth to complete a full revolution around the Sun.
(a)(iii) 24 hours. This is the period of one full rotation of the Earth on its internal axis, resulting in the standard day-night cycle observed from the surface.
(a)(iv) Hydrogen and helium. The Sun is composed primarily of these two light elements, with hydrogen undergoing nuclear fusion in the core to produce helium and vast amounts of energy.
(b)(i) Infrared or ultraviolet. In addition to visible light, the Sun emits significant radiation in the infrared (heat) and ultraviolet regions of the electromagnetic spectrum.
(b)(ii) 500 s. Using the wave speed equation t= v d , we divide the Earth-Sun distance (1.5×10 11 m) by the speed of light (3.0×10 8 m/s) to get 500 seconds (approximately 8.3 minutes).