Question 1

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▶️ Answer/Explanation
Detailed solution:
Scalar quantities have only magnitude, while vector quantities possess both magnitude and direction.
Acceleration is a vector because it is the rate of change of velocity, which inherently includes direction.
Force is a vector as it requires both magnitude and direction for its complete description (e.g., 10 N downwards).
Momentum depends on velocity, so it is also a vector quantity aligned with the direction of motion.
Option A correctly lists acceleration, force, and momentum all under the vector quantities column.
The other options incorrectly place at least one vector quantity in the scalar column.
Questions 2
A. density
B. mass
C. volume
D. weight
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Mass is defined as the measure of the quantity of matter contained in an object at rest relative to the observer.
Weight is a gravitational force acting on that mass, while density is mass per unit volume and volume is the space occupied.
Since the question asks specifically for the term describing the “quantity of matter,” mass is the correct scientific term.
Unlike weight, mass does not change based on location or gravitational field strength.
Therefore, option B correctly identifies the quantity of matter in an object.
Question 3

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Mass is the measure of the quantity of matter in a body and remains constant regardless of location.
Weight is the gravitational force acting on a mass, given by W = mg, so it depends on the gravitational field strength (g).
Moving the body to a place with a different g changes the gravitational pull, thus altering its weight.
Since mass does not change but weight does, option C correctly identifies mass as unchanged and weight as changed.
This distinction is fundamental in physics, especially when comparing measurements on Earth versus the Moon.
Therefore, option C is the correct answer.
Question 4
A. \(5.2 \times 10^{–3} g / cm^3\)
B. \(5.2g / cm^3\)
C. \(190 g / cm^3\)
D. \(1.2 \times 10^3 g / cm^3\)
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
First convert mass to grams: \(2.5 \text{ kg} = 2500 \text{ g}\). Then use \(\rho = \frac{m}{V} = \frac{2500}{480} \approx 5.2 \text{ g/cm}^3\).
The formula defines density as mass per unit volume, requiring consistent units for calculation.
Option A is incorrect due to a decimal error, while C and D result from faulty arithmetic or unit conversion.
Thus, the correct density is \(5.2 \text{ g/cm}^3\), matching option B.
Question 5

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
An object is in equilibrium when the resultant force acting on it is zero, meaning forces are balanced in all directions.
In option A, the two forces are equal in magnitude (4.0 N) and opposite in direction, so they cancel each other out.
In options B, C, and D, the opposing forces have different magnitudes, resulting in a non-zero net force and acceleration.
Since only object A has no resultant force, it remains at rest or moves with constant velocity according to Newton’s first law.
Therefore, option A correctly identifies the object in equilibrium.
Question 6
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Momentum is a vector, so direction matters. Taking downwards as positive, initial momentum = \(+mv_1\).
After rebound, velocity is upwards, so final momentum = \(-mv_2\).
Change in momentum = final momentum − initial momentum = \((-mv_2) – (+mv_1) = -m(v_1 + v_2)\).
The magnitude of the change in momentum is therefore \(m(v_1 + v_2)\).
Option C correctly represents this magnitude, accounting for the reversal in direction.
Question 7
A. 17 m/s
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Using \(v^2 = u^2 + 2as\) with \(u=0\), \(a=9.8\,\mathrm{m/s^2}\), and \(s=50\,\mathrm{m}\), we find \(v = \sqrt{2 \times 9.8 \times 50} = \sqrt{980} \approx 31.3\,\mathrm{m/s}\).
The mass of the stone (3.0 kg) is not needed because acceleration due to gravity is independent of mass.
Rounding to two significant figures gives approximately \(31\,\mathrm{m/s}\).
Thus, the stone hits the ground at a speed of \(31\,\mathrm{m/s}\).
Question 8

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Coal is a fossil fuel that stores chemical energy, which is released when it is burned.
The moving generator possesses kinetic energy due to its motion as it rotates to produce electricity.
Energy is transferred from the chemical store in the coal to the kinetic store of the generator via thermal and mechanical pathways.
The other options incorrectly identify either the coal’s store as internal/thermal or the generator’s store as chemical.
Therefore, row B correctly identifies the energy stores as chemical for coal and kinetic for the moving generator.
Question 9

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The ice turning into liquid water inside the glass is an example of melting (solid to liquid).
The water droplets forming on the outside of the cold glass occur because water vapor in the warm air cools upon contact with the surface and turns into liquid water, a process known as condensation.
Therefore, the two changes of state taking place are condensation and melting.
Question 10
Gas of volume 200 \(cm^3\) is trapped inside a container by a piston. The piston is pushed to the right and the volume of the gas decreases to 100 \(cm^3\). The temperature of the gas remains constant.

Which row states the effect that this has on the kinetic energy of the gas particles and the force per unit area exerted by the particles colliding with the inside walls of the container? 
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Since the temperature remains constant, the average kinetic energy of the gas particles does not change.
Decreasing the volume forces the same number of particles into a smaller space, increasing the frequency of collisions with the walls.
This results in a greater force per unit area, meaning the pressure exerted by the gas increases.
Option C correctly identifies that the kinetic energy stays the same while the force per unit area increases.
Therefore, the relationship between volume and pressure at constant temperature is demonstrated.
Question 11

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
For a fixed mass of gas at constant temperature, Boyle’s law states that pressure (\(p\)) is inversely proportional to volume (\(V\)), i.e., \(pV =\) constant. This relationship produces a hyperbolic curve on a \(p\)–\(V\) graph, where pressure decreases as volume increases. Graphs A and B show direct proportions, graph C shows constant pressure, but only graph D correctly illustrates the inverse relationship. The curve in D approaches both axes asymptotically, confirming that \(p \propto 1/V\). Therefore, option D is the correct representation of the pressure–volume relationship at constant temperature.
Question 12

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Evaporation occurs at the surface of a liquid, not throughout its bulk.
The most energetic particles escape, reducing the average kinetic energy of the remaining particles.
Since temperature is a measure of average kinetic energy, the liquid cools down.
Therefore, the correct row identifies escape from the surface with a temperature decrease.
This matches option C in the provided table.
Question 13
A. a cold day with no wind
B. a cold day with wind
C. a hot day with no wind
D. a hot day with wind
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Evaporation occurs when the most energetic particles escape from the surface of a liquid.
A higher temperature increases the average kinetic energy of particles, speeding up evaporation.
Wind removes the water vapor from the air above the clothes, maintaining a low concentration gradient and allowing more particles to escape.
Therefore, a hot day with wind combines the two most significant factors (temperature and air movement) that increase the rate of evaporation.
A cold day reduces particle energy, while still air allows the air to become saturated with vapor, both of which slow the process.
Thus, option D provides the conditions for the fastest rate of evaporation.
Question 15
A. dull black
B. shiny black
C. dull white
D. shiny white
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Dull, black surfaces are the best absorbers and emitters of infrared radiation.
A solar water heater needs to absorb as much thermal energy from sunlight as possible.
Shiny or white surfaces are good reflectors and poor absorbers, so they would be less efficient.
The dull texture prevents reflection, while the black colour maximises absorption of incident radiation.
Therefore, a dull black surface is the optimal choice for heating water efficiently using solar energy.
Question 16
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Waves are disturbances that propagate through a medium or space, carrying energy from one location to another.
As a wave travels, particles of the medium oscillate about their equilibrium positions but do not travel with the wave.
This means the net displacement of matter is zero, while the energy of the disturbance moves forward.
Options A and D are incorrect because waves definitely transfer energy; option B is incorrect because matter is not transferred.
Examples include water waves moving energy across a pond while a floating object simply bobs up and down.
Thus, the fundamental definition of a wave is confirmed by option C.
Question 17
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
According to the law of reflection, the angle of incidence \(i\) equals the angle of reflection \(r\).
The angle between the incident and reflected rays is the sum \(i + r = 2i\).
As the angle of incidence increases toward grazing incidence, \(i\) approaches \(90^\circ\).
Therefore, the maximum angle between the two rays approaches \(2 \times 90^\circ = 180^\circ\).
This corresponds to the incident ray striking the mirror almost parallel to its surface.
Hence, option D is correct.
Question 18

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▶️ Answer/Explanation
Detailed solution:
Total internal reflection occurs when light travels from a denser medium to a less dense medium, and the angle of incidence exceeds the critical angle.
If the angle is smaller than the critical angle, the light will refract out of the denser medium rather than reflect entirely.
The table in option D correctly indicates that the angle must be greater than the critical angle for total internal reflection to happen.
This principle is essential in optical fibres and certain prism applications in physics.
Therefore, option D correctly identifies both necessary conditions for this phenomenon.
Question 19

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The refractive index \(n\) is calculated using Snell’s Law: \(n = \frac{\sin i}{\sin r}\).
From the diagram, the angle of incidence \(i\) in air is \(40^\circ\) and the angle of refraction \(r\) in the substance is \(30^\circ\).
Substituting the values: \(n = \frac{\sin 40^\circ}{\sin 30^\circ} = \frac{0.6428}{0.5} \approx 1.2856\).
Rounding to three significant figures gives \(1.29\), which matches option B.
This aligns with the syllabus definition of refractive index as the ratio of the sine of the angle of incidence to the sine of the angle of refraction.
Question 20
The diagram shows a ray of light in an optical fibre.

Which statement correctly explains the condition for the maximum transmission of light by the optical fibre?
A. The glass must slow the light as little as possible to make the critical angle for the fibre as large as possible.
B. The glass must slow the light as little as possible to make the critical angle for the fibre as small as possible.
C. The glass must slow the light as much as possible to make the critical angle for the fibre as large as possible.
D. The glass must slow the light as much as possible to make the critical angle for the fibre as small as possible.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
For maximum transmission, light must undergo total internal reflection; this requires the angle of incidence to exceed the critical angle \( c \).
Since \( n = \frac{1}{\sin c} \), a higher refractive index (slower light in glass) results in a smaller critical angle, making total internal reflection more likely.
Therefore, the glass must slow light as much as possible to make the critical angle as small as possible.
Question 21
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Ultraviolet (UV) radiation is commonly used in security marking to detect counterfeit currency.
Many bank notes contain special fluorescent inks or security threads that glow under UV light but remain invisible in normal lighting.
Radio waves and microwaves are used for communication and heating, not for detecting security features on paper.
X-rays have much higher energy and are used for medical imaging and security scanning of objects, not bank notes.
The fluorescence effect caused by UV light allows quick and reliable authentication of genuine bank notes.
Therefore, option C correctly identifies ultraviolet as the region used for detecting fake bank notes.
Question 22
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The distance from a compression to the nearest rarefaction is half a wavelength (λ/2).
Given λ/2 = 2.5 cm = 0.025 m, the wavelength λ = 0.05 m.
Using the wave equation v = f λ, with v = 330 m/s, we find f = v / λ = 330 / 0.05 = 6600 Hz.
Thus, the frequency of the sound wave is 6600 Hz, corresponding to option C.
Question 23
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▶️ Answer/Explanation
Detailed solution:
Ultrasound refers to sound waves with frequencies above the human hearing range (greater than 20 kHz).
These high-frequency waves can penetrate soft tissue and reflect off internal structures to create diagnostic images.
Gamma rays are ionizing radiation used for cancer treatment or sterilization, not primarily for soft tissue scanning.
Infrared is associated with thermal imaging, and microwaves are used for communications and cooking.
Therefore, due to its non-ionizing nature and ability to image soft organs, ultrasound is the correct answer.
Question 24

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▶️ Answer/Explanation
Detailed solution:
A temporary magnet, such as an electromagnet core, is made of soft iron because it gains and loses magnetism easily.
A permanent magnet is made of steel because it retains its magnetism over a long period of time.
This distinction is fundamental to the applications of magnets in devices like relays and compasses.
Therefore, soft iron is suitable for a temporary magnet and steel is suitable for a permanent magnet.
Question 25

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▶️ Answer/Explanation
Detailed solution:
The strength of a magnetic field is indicated by the concentration or spacing of the magnetic field lines.
Where the field lines are closest together, the magnetic field is strongest.
In the diagram, the field lines are most densely packed at point A, near the pole of the magnet.
As the distance from the pole increases (points B, C, and D), the lines spread further apart, indicating a weaker field.
Therefore, the magnetic field is strongest at point A.
Question 26

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
To measure resistance using a voltmeter and an ammeter, the voltmeter must be connected in parallel with the resistor to measure potential difference across it, while the ammeter must be in series to measure the current through it. In circuit D, the voltmeter is correctly placed across resistor R, and the ammeter is in series with R, allowing accurate readings for the calculation R = V/I. Circuits A and B place the voltmeter in series or the ammeter in parallel, which is incorrect. Circuit C measures total current but not the correct voltage across R alone. Therefore, circuit D shows the proper arrangement for determining resistance.
Question 27
A diode and a resistor are connected across a variable d.c. supply. Terminal X is initially positive and at a maximum value. The potential difference (p.d.) across XY is adjusted so that it changes, as shown by the graph.

Which graph shows how the current in the circuit (from X to Y) varies during the same time interval? 
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
A diode allows current to flow only when it is forward biased (X positive relative to Y).
The given p.d. graph shows X positive initially, so current flows; as p.d. decreases, current decreases.
When the p.d. becomes negative (Y positive relative to X), the diode is reverse biased and blocks current.
Therefore, the current graph must show a positive pulse followed by zero current, matching the p.d. shape.
Option D correctly shows current only during the positive half of the p.d. variation and zero otherwise.
Hence, the current follows the forward bias condition of the diode, making D the correct graph.
Question 28
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
According to Lenz’s law, the direction of the induced current is always such that it opposes the change in magnetic field that produced it.
This principle is a consequence of the conservation of energy; if the induced current aided the change, it would create a perpetual motion scenario.
Therefore, the induced magnetic field created by the current will try to repel the approaching magnet or attract the departing magnet.
Options A, B, and C are incorrect as they do not account for the specific opposition to the change in flux linkage.
Hence, the correct statement is that the direction of the induced current opposes the change causing it.
Question 29


Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The magnetic field inside a solenoid (at point X) is strong and uniform due to the concentrated field lines.
Outside the solenoid, the field strength decreases significantly; point Y, being near the coils, has a weaker field than X.
Point Z is farther away from the solenoid, where the field is very weak, essentially negligible compared to X.
Thus, the field strength at Y is weaker than at X, and at Z it is much weaker than at X.
This matches the comparison given in option A of the table.
Therefore, option A is correct.
Question 30
A conducting wire is placed between the poles of a magnet. When an electric current in the wire is in the direction shown, then the force on the wire acts out of the page.

Three statements of different conditions and how the wire is affected are given.
1 When the current is towards the top of the page and the direction of the magnetic field is unchanged, the force produced acts into the page.
2 When the current is towards the bottom of the page and the magnetic field is reversed, the force produced acts into the page.
3 When the current in the wire is alternating, the wire vibrates into and out of the page.
Which statements are correct?
A. 1, 2 and 3
B. 1 and 2 only
C. 1 and 3 only
D. 2 and 3 only
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
According to Fleming’s left-hand rule, reversing the current (statement 1) reverses the force direction to into the page.
Reversing both the current and the magnetic field (statement 2) results in the same original force direction, which is into the page.
An alternating current (statement 3) continuously changes direction, causing the force to alternate between into and out of the page, making the wire vibrate.
Since all three statements correctly describe the outcomes based on electromagnetic principles, statements 1, 2, and 3 are all correct.
Question 31

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Transformers operate on the principle of electromagnetic induction, which requires a changing magnetic field.
A direct current (d.c.) produces a steady magnetic field, so it cannot induce an e.m.f. in the secondary coil.
The alternating current (a.c.) in the primary coil creates a continuously changing magnetic flux in the core.
This changing flux links with the secondary coil and induces an alternating current (a.c.) of a different voltage.
Therefore, both the primary input and the secondary output must be alternating current (a.c.) for the transformer to function.
Question 32
A. Alpha-particles are attracted to the nucleus of the atoms in the metal sheet.
B. Half the mass of the atom is concentrated in the nucleus.
C. Most of the alpha-particles are deflected, showing that the nucleus is very dense.
D. Only a small number of alpha-particles are deflected, but some of these are deflected through large angles.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
In Rutherford’s gold foil experiment, most alpha-particles passed straight through, indicating the atom is mostly empty space.
Only a very small number were deflected, with a tiny fraction bouncing back at large angles, proving the existence of a dense, positively charged nucleus.
Option A is incorrect as alpha-particles are positively charged and are repelled by the nucleus, not attracted to it.
Option B is false because almost all the mass is concentrated in the tiny nucleus, not just half.
Option C is wrong as most particles are not deflected; the few large deflections reveal the nucleus’s density, not the majority path.
Thus, D accurately summarizes the key experimental observation supporting the nuclear model.
Question 33

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▶️ Answer/Explanation
Detailed solution:
The notation shows the nucleon number (top) is 197 and the proton number (bottom) is 79.
The proton number (Z) indicates the number of protons in the nucleus, which determines the element.
Since each proton carries a relative charge of +1 and neutrons have zero charge, the total nuclear charge depends only on protons.
Therefore, the relative charge on the gold nucleus is equal to its proton number, which is +79.
Options B, C, and D correspond to neutron number, nucleon number, and their sum, respectively, and are incorrect.
Thus, the correct answer is A.
Question 34

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▶️ Answer/Explanation
Detailed solution:
The nuclide notation \(_{1}^{2}\textrm{H}\) shows a nucleon number \(A=2\) and a proton number \(Z=1\).
The number of neutrons is calculated as \(A – Z = 2 – 1 = 1\) neutron.
The relative charge on the nucleus is determined solely by the protons; since \(Z=1\), the charge is \(+1\).
Neutrons contribute mass but have zero charge, so they do not affect the overall nuclear charge.
Therefore, the nucleus contains 1 neutron and has a relative charge of \(+1\), corresponding to row A in the table.
Question 35

A. \(\alpha\)-particles and \(\gamma\)-rays
B. \(\alpha\)-particles only
C. \(\beta\)-particles and \(\gamma\)-rays
D. \(\beta\)-particles only
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Paper alone reduces the count significantly but not completely, showing that alpha particles (stopped by paper) are not the only radiation present.
The 5 mm aluminium sheet further reduces the count but does not block all radiation, indicating the presence of beta particles which are absorbed by a few millimetres of aluminium.
Since some radiation still passes through the aluminium, it must be gamma rays, which are highly penetrating and require thick lead to be stopped.
The source therefore emits both beta particles (stopped by aluminium) and gamma rays (penetrating both materials).
Alpha particles are not emitted because paper would completely stop them, leaving no further reduction with aluminium.
Question 36
A. 44 counts per second
B. 123 counts per second
C. 134 counts per second
D. 167 counts per second
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The initial count rate from the source is 536 – 44 = 492 counts/s.
After 68 hours, which is two half-lives (68 ÷ 34 = 2), the source count rate halves twice to 492 ÷ 4 = 123 counts/s.
The detector reading is the sum of the remaining source count rate and the constant background count rate.
Therefore, the total reading is 123 + 44 = 167 counts per second.
This corresponds to option D.
Question 37

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
When a planet is closest to the Sun in its elliptical orbit, it has the greatest orbital speed due to conservation of energy.
The gravitational potential energy is at a minimum here, meaning kinetic energy and speed are at a maximum.
Conversely, when farthest from the Sun, the speed is at a minimum.
This principle applies to all objects in elliptical orbits, including planets and comets.
Row B correctly identifies the maximum speed occurring at the point closest to the Sun.
Therefore, option B accurately describes the relationship between orbital distance and speed.
Question 38
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The Sun generates energy through nuclear fusion in its core, not chemical burning or fission.
In this process, hydrogen nuclei combine under extreme pressure and temperature to form helium.
A small amount of mass is converted into a large amount of energy during this reaction.
Fission is the splitting of heavy nuclei, which is not the primary process occurring in the Sun.
Therefore, the correct description of the Sun’s power source is the nuclear fusion of hydrogen.
Question 39
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Using Hubble’s Law \(v = H_0 \times d\) with \(H_0 \approx 2.2 \times 10^{-18} \text{ s}^{-1}\).
Convert distance to metres: \(d = 3.0 \times 10^{20} \text{ km} = 3.0 \times 10^{23} \text{ m}\).
Calculate recessional speed: \(v = (2.2 \times 10^{-18}) \times (3.0 \times 10^{23}) = 6.6 \times 10^5 \text{ m/s} = 660 \text{ km/s}\).
This matches option A, confirming the galaxy’s recessional velocity is 660 km/s.
Question 40

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Distance to a far galaxy is determined using the brightness of a supernova, as supernovae have a known peak luminosity.
The speed of recession is found from the change in observed wavelength (redshift) of the galaxy’s light.
Redshift measures how much the wavelength has stretched due to the expansion of the Universe, giving the velocity.
This method aligns with Hubble’s Law and the use of standard candles for cosmic distance measurements.
Therefore, row A correctly pairs the method of determining distance with the method of determining speed.
