Question 1
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The object travels at constant speed between 5 s and 15 s, so the time interval is $10 \text{ s}$.
The speed during this interval is $5 \text{ m/s}$ (from the graph’s horizontal section).
Distance travelled = speed × time = $5 \text{ m/s} \times 10 \text{ s} = 50 \text{ m}$.
This corresponds to the area under the graph for that section (a rectangle of height 5 and width 10).
Hence, the correct answer is Option B.
Question 2
Which statement about a falling object accelerating close to the Earth’s surface is correct?
A. The weight of the object is increasing and the force of air resistance on the object is decreasing.
B. The weight of the object and the force of air resistance on the object are of equal magnitude, but act in opposite directions.
C. The weight of the object is constant, but the force of air resistance on the object is increasing.
D. The weight of the object is less than the force of air resistance.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Weight ($W=mg$) remains constant as mass and gravitational field strength do not change.
As speed increases, air resistance (drag) increases progressively.
The object accelerates until the increasing air resistance equals the constant weight, at which point it reaches terminal velocity.
Therefore, during acceleration, weight is constant while air resistance is increasing.
This matches Option C.
Question 3
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Velocity is a vector, so we must combine the perpendicular components using vector addition.
The aircraft’s velocity north and the wind’s velocity east form a right-angled triangle.
Resultant magnitude = √(60² + 13²) = √(3600 + 169) = √3769.
Calculating the square root gives approximately 61.4 m/s, which rounds to 61 m/s.
Therefore, the magnitude of the resultant velocity is 61 m/s.
Question 4
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Mass is the quantity of matter in an object, independent of gravitational field strength.
A beam balance compares masses directly by balancing gravitational forces, which cancel out in the equation.
Hanging from a spring measures weight (force), which depends on gravity, not mass directly.
Therefore, only methods that use balancing principles (like a beam balance) provide a true comparison of mass.
This confirms that Options 3 and 4 are the correct methods.
Question 5
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Mass is constant everywhere; calculate it using the probe data: $m = W / g = 3.5\text{ N} / 7.0\text{ N/kg} = 0.50\text{ kg}$.
Then find the weight on Earth’s surface using $W = mg = 0.50\text{ kg} \times 9.8\text{ N/kg} = 4.9\text{ N}$.
The mass remains $0.50\text{ kg}$ and the new weight is $4.9\text{ N}$, which matches Option B.
Question 6
A cyclist is travelling in a straight line along a horizontal road at a constant speed. A constant driving force F acts on the cyclist in the forward direction shown. Which statement about the magnitude of the frictional forces acting on the cyclist is correct?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Since the cyclist is moving at a constant speed in a straight line, the resultant force is zero according to Newton’s first law.
This means the forward driving force $F$ must be exactly balanced by the total frictional forces (including air resistance) acting in the opposite direction.
For the forces to balance, the magnitude of the frictional forces must be equal to $F$.
Therefore, option A is correct; the other options would imply a non-zero resultant force, causing acceleration or deceleration.
Question 7
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The spring constant $k$ is defined as the force per unit extension, given by $k = \frac{F}{x}$.
The extension $x$ is the change in length: $5.0\text{ cm} – 3.0\text{ cm} = 2.0\text{ cm}$.
Substituting the values: $k = \frac{60\text{ N}}{2.0\text{ cm}} = 30\text{ N/cm}$.
Since the limit of proportionality is not exceeded, the spring obeys Hooke’s law.
This matches Option D.
Question 8
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
According to the principle of moments, Force × Perpendicular distance from pivot = Load × Perpendicular distance from pivot.
To reduce the required force (F2), the perpendicular distance from the force to the pivot must be increased.
Situation P increases the length of the lever, increasing the force’s distance from the pivot. Situation Q moves the pivot away from the log, increasing the force’s distance.
Situation R moves the pivot closer to the log, which decreases the force’s distance, thereby increasing the required force.
Therefore, F2 is smaller than F1 only in situations P and Q.
Question 9
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Impulse is defined as the change in momentum: $Impulse = \Delta p = m(v – u)$.
Taking the initial direction as positive, $u = 16\text{ m/s}$ and $v = -12\text{ m/s}$.
The change in velocity is $16 – (-12) = 28\text{ m/s}$.
Thus, $Impulse = 0.25\text{ kg} \times 28\text{ m/s} = 7.0\text{ N s}$.
This matches Option D.
Question 10
A bicycle braking system transfers energy from a kinetic energy store to an internal energy store. A motor converts energy from a chemical energy store (battery) to a kinetic energy store. What enables these energy transfers?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The bicycle braking system uses friction, which is a force, so energy is transferred by mechanical work.
The motor uses an electric current from the battery, so energy is transferred by electrical work.
This matches the descriptions for energy transfers by forces and electrical currents in the syllabus.
Therefore, the correct combination is mechanical work for the braking system and electrical work for the motor.
Question 11
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• TOPIC 5.1.2: The nucleus (Describe the processes of nuclear fusion as the joining of nuclei)
▶️ Answer/Explanation
Detailed solution:
Fusion requires extremely high temperatures to give nuclei enough kinetic energy to overcome the strong electrostatic repulsion.
Hydrogen nuclei (protons) are positively charged, so they naturally repel each other, making it difficult to achieve the high density required for fusion.
The high temperature is needed to create a plasma state, while the repulsion between positive charges hinders confinement.
This confirms that the temperature needed is very high and the difficulty arises from the repulsion of positively charged nuclei.
Therefore, Option D accurately identifies both problems.
Question 12
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Power is defined as the rate of doing work or the rate of energy transfer, given by $P = \frac{\Delta E}{t}$.
A large power output does not strictly require a large force, a fast speed, or a large engine volume.
It specifically means that a significant amount of energy is being transferred or converted per unit time.
Therefore, a vehicle engine with large power must transfer large amounts of energy each second.
This makes Option C the only necessarily correct statement based on the definition of power.
Question 13
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Using the formula for liquid pressure $P = \rho g h$, rearrange to find density: $\rho = \frac{P}{g h}$.
From the graph, at a depth $h = 0.4\text{ m}$, the pressure $P = 8000\text{ Pa}$.
Substitute values: $\rho = \frac{8000}{9.8 \times 0.4} = \frac{8000}{3.92} \approx 2040.8\text{ kg/m}^3$.
This is closest to $2000\text{ kg/m}^3$, matching Option C.
The calculation confirms the liquid’s density is approximately $2000\text{ kg/m}^3$.
Question 14
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Absolute zero is defined as the lowest possible temperature, which corresponds to 0 K on the Kelvin scale or –273 °C on the Celsius scale.
At this theoretical temperature, the motion of particles is at a minimum, meaning they possess the least possible kinetic energy (though they retain some quantum zero-point energy).
Gravitational potential energy depends on mass and position in a field, not on thermal energy, so it is unaffected by reaching absolute zero.
Therefore, the statement in Option A correctly identifies both the temperature value and the state of particle energy.
Question 15
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Condensation is the change of state from a gas to a liquid.
In a liquid, particles are arranged close together but are not in fixed positions; they can slide over one another.
This distinguishes them from solids, where particles vibrate about fixed points.
Therefore, the correct description of particles after condensation is that they are close together and slide over each other.
This matches Option A.
Question 16
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Black surfaces are better absorbers and better emitters of infrared radiation than white or shiny surfaces.
Therefore, in sunshine, the black car absorbs radiation faster and heats up more quickly.
At night, the black car emits its internal thermal energy as radiation faster, cooling down more rapidly.
Hence, the black car shows the greater rate of both temperature increase and temperature decrease.
This aligns with the principles of thermal radiation covered in the syllabus.
Question 17
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Water waves on the surface are transverse because the water moves up and down while the wave travels horizontally.
Since the water depth is constant, the medium is uniform and the waves spread out equally from the point of impact.
Therefore, the wave speed remains the same in all directions.
This matches Option C.
Question 18
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
A magnifying glass requires a virtual, upright, and magnified image. For a converging lens, this occurs only when the object is placed inside the focal length (between the optical center and point F).
In this position, the refracted rays diverge and do not meet on the opposite side; instead, they appear to come from a larger, upright virtual image formed on the same side as the object.
Options A and D describe real image formation, and Option B describes an image at infinity, none of which function as a practical magnifying glass.
Question 19
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
When light travels from one medium to another, its speed and wavelength change due to the change in optical density.
However, the frequency of the light is determined by the source and remains constant during refraction.
Since the light is monochromatic (single frequency), this property is unchanged as it enters the glass.
Therefore, the frequency always remains the same, making Option C the correct choice.
Question 20
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
When white light undergoes dispersion in a prism, red light is deviated the least while blue/violet light is deviated the most.
The spectrum always appears with red at one end and blue/violet at the other, with yellow in between.
Therefore, if the colours appear in sequence from least to most deviated, the correct order is red, yellow, blue.
This corresponds to Option C.
Question 21
Which statement is correct?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Wireless internet and mobile phones typically use microwaves or radio waves, not visible light or X-rays.
Bluetooth uses radio waves, not ultraviolet.
Cable television and high-speed broadband often use optical fibres, which transmit data using visible light or infrared radiation.
Since glass is transparent to infrared, this is a common and correct application.
Therefore, statement C correctly identifies a use of infrared wavelengths in communication.
Question 22
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The sound travels from the ship to the cliff and back, so the total distance covered is $2 \times 790\text{ m} = 1580\text{ m}$.
Using the formula $\text{speed} = \frac{\text{distance}}{\text{time}}$, we calculate $\frac{1580\text{ m}}{4.8\text{ s}} \approx 329.2\text{ m/s}$.
Rounding to two significant figures gives $330\text{ m/s}$, which is the typical speed of sound in air.
This calculation demonstrates a standard method for measuring the speed of sound using an echo.
Thus, option B is the correct answer.
Question 23
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Electromagnets use a soft iron core because iron is easily magnetized and demagnetized.
When current flows, the iron becomes strongly magnetized; when current stops, it loses magnetism.
This temporary magnetic property is essential for controlled applications like relays and bells.
Steel retains magnetism and is therefore used for permanent magnets, not electromagnets.
Option B correctly identifies the material (iron) and the resulting property (temporary magnet).
Question 24
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
When the rod becomes negatively charged, it has gained negatively charged electrons from the cloth.
Since electrons moved from the cloth to the rod, the cloth lost negative charge and is left with a positive charge.
Neutrons are neutral and do not move during electrostatic charging.
Therefore, the charge on the cloth is positive and the particles that moved are electrons.
This matches Option C.
Question 25
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
For a metallic wire at constant temperature, resistance is directly proportional to its length.
Therefore, in an experiment investigating this relationship, the length of the wire is the independent variable and is plotted on the x-axis.
The resistance, being the dependent variable, would be plotted on the y-axis, producing a straight line through the origin.
This is a standard method to verify the proportional relationship between resistance and length.
Thus, the correct option is B.
Question 26
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Convert power to kilowatts: $2200 \text{ W} = 2.2 \text{ kW}$.
Convert time to hours: $48 \text{ minutes} = \frac{48}{60} \text{ hours} = 0.8 \text{ hours}$.
Calculate energy used: $E = P \times t = 2.2 \text{ kW} \times 0.8 \text{ h} = 1.76 \text{ kWh}$.
Calculate total cost: $\text{Cost} = 1.76 \text{ kWh} \times \$0.25/\text{kWh} = \$0.44$.
This matches Option A.
Question 27
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Resistance \( R \) is directly proportional to length \( l \) and inversely proportional to cross-sectional area \( A \) (where \( A \propto d^2 \)).
Wire B has the shortest length (2.0 m) and the largest diameter (1.5 mm), giving it the largest area.
Short length minimizes \( R \), and large area further reduces \( R \), so Wire B has the smallest resistance.
Comparing dimensions: B is shorter than C and D, and thicker than A and C.
Thus, Option B is correct.
Question 28
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
To light only lamps P and R, current must flow through their specific branches while bypassing Q and S.
Closing switch 1 connects the power source to the junction leading to P and R, and closing switch 3 completes the path for P and R while keeping the branch containing Q and S open (switch 2 open).
This configuration ensures that only the desired lamps are part of a complete circuit.
Therefore, only switches 1 and 3 need to be closed, matching Option C.
Question 29
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Increasing the current flowing through a solenoid produces a stronger magnetic field around it.
The spacing of magnetic field lines indicates the field’s strength; closer lines represent a stronger field.
Therefore, as the current increases, the field strength increases and the field lines become more concentrated (closer together).
This direct relationship matches the description provided in Option C.
Question 30
The diagram shows a wire hanging freely between the poles of a magnet. There is a current in the wire in the direction shown. The magnet and current cause a force to act on the wire. In which direction does this force act?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Fleming’s left-hand rule gives the direction of force on a current-carrying conductor in a magnetic field.
Align the First finger with the magnetic Field (N to S), the seCond finger with the Current, and the Thumb shows the direction of the Thrust (Force).
Applying this rule to the described setup, the force is directed out of the plane of the page, toward the viewer.
This matches Option B: out of the page (toward you).
Question 31
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
A simple a.c. generator uses slip rings to maintain continuous electrical contact, allowing current to flow in alternating directions.
A d.c. motor, however, requires a split-ring commutator to reverse the current direction every half turn.
This reversal ensures the force on the coil remains in the same rotational direction, allowing continuous rotation.
The coil, brushes, and magnet are common to both devices.
Therefore, the split-ring commutator is unique to the d.c. motor.
Question 32
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
First, use the turns ratio equation $\frac{V_{\mathrm{p}}}{V_{\mathrm{s}}} = \frac{N_{\mathrm{p}}}{N_{\mathrm{s}}}$ to find the primary voltage: $\frac{V_{\mathrm{p}}}{110} = \frac{5500}{500}$, giving $V_{\mathrm{p}} = 1210\text{ V}$.
Since the transformer is $100\%$ efficient, input power equals output power: $P_{\mathrm{p}} = P_{\mathrm{s}} = 132\text{ W}$.
Use the power equation $P = IV$ for the primary coil: $I_{\mathrm{p}} = \frac{P_{\mathrm{p}}}{V_{\mathrm{p}}} = \frac{132}{1210} = 0.109\text{ A} \approx 0.11\text{ A}$.
Wait, $132 / 1210 = 0.10909…$, which rounds to $0.11\text{ A}$, not $0.12\text{ A}$.
However, the question provides the output voltage directly as $110\text{ V}$. For an ideal transformer, $I_{\mathrm{p}}V_{\mathrm{p}} = I_{\mathrm{s}}V_{\mathrm{s}}$. The secondary current $I_{\mathrm{s}} = \frac{132}{110} = 1.2\text{ A}$. Then $I_{\mathrm{p}} = I_{\mathrm{s}} \times \frac{N_{\mathrm{s}}}{N_{\mathrm{p}}} = 1.2 \times \frac{500}{5500} = 0.109\text{ A}$.
There appears to be a rounding difference in the provided answer. The calculation yields $0.11\text{ A}$, which corresponds to Option A, but the provided answer indicates Option B ($0.12\text{ A}$). The detailed solution above matches the logic but notes the discrepancy.
Question 33
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The scattering of alpha particles showed that most passed through (empty space) while a few were deflected at large angles (small, dense, positively charged nucleus).
This experiment demonstrated the concentration of mass and charge in the nucleus but did not provide evidence for the existence of neutrons.
The discovery of neutrons came later through the work of James Chadwick.
Therefore, the conclusion that the nucleus consists of protons and neutrons is not derived from Rutherford’s gold foil experiment.
Question 34
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The nucleon number (mass number) is represented by the superscript $A$ in the nuclide notation $^{A}_{Z}X$.
It is the sum of protons and neutrons in the nucleus; for this chlorine nuclide, the value is $35$.
The subscript $17$ is the proton number (atomic number), not the nucleon number.
Therefore, the nucleon number is $35$, which corresponds to Option C.
Question 35
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
In β⁻ decay, a neutron-rich nucleus undergoes a transformation where a neutron changes into a proton.
This process emits an electron (the β-particle) and an antineutrino to conserve charge and energy.
The proton remains in the nucleus, increasing the atomic number by 1 while the mass number stays the same.
Therefore, the correct nuclear change is: neutron → proton + electron.
Question 36
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
First, correct for background radiation: subtract 5 counts/min from all readings.
The initial corrected count rate is therefore $45 – 5 = 40$ counts/min.
Half of this corrected value is $20$ counts/min, which corresponds to a measured reading of $25$ counts/min on the graph.
Reading from the time axis, the count rate falls from $45$ to $25$ counts/min over a period of $2.5$ days.
Thus, the half-life of the isotope is $2.5$ days, matching Option B.
Question 37
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The Earth completes one full orbit around the Sun in approximately $365$ days, defining our calendar year.
The Moon takes roughly one month (about $30$ days) to complete its orbit around the Earth.
This matches the values given in row D, while the other options incorrectly swap or misstate these periods.
Therefore, the combination of $365$ days for Earth’s orbit and $30$ days for the Moon’s orbit is correct.
Question 38
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Comets travel around the Sun in highly elongated elliptical paths, not perfect circles.
In an elliptical orbit, the Sun is located at one of the two foci of the ellipse, which is offset from the geometric center.
This explains why comets spend most of their time far from the Sun and move very quickly when they swing close around it.
Option B correctly identifies the shape as elliptical and the Sun’s position as off-center.
Question 39
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
In a stable star, the immense gravitational pressure creates conditions for nuclear fusion in the core.
During this process, hydrogen nuclei (protons) combine to form helium, releasing vast amounts of energy.
This outward thermal pressure balances the inward pull of gravity, maintaining the star’s stability.
Fission (splitting heavy nuclei like uranium) is not the primary process powering stable stars.
Therefore, the correct description is nuclear fusion using hydrogen as fuel, matching Option C.
Question 40
Which quantity can be determined using the brightness of a supernova in a distant galaxy?
A. the speed at which the galaxy is moving away from the Earth
B. the distance of the galaxy from the Earth
C. the Hubble constant
D. the age of the Universe
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Type Ia supernovae have a known peak intrinsic luminosity, making them standard candles.
By comparing their known absolute brightness to their observed apparent brightness from Earth, astronomers can calculate the distance to the host galaxy.
This method provides a crucial rung on the cosmic distance ladder for measuring far-away galaxies.
Therefore, the brightness of a supernova is used specifically to determine its distance.
This matches Option B.
