Question 1

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The two forces of $3\text{ N}$ and $4\text{ N}$ are perpendicular, so the magnitude of the resultant is found using Pythagoras’ theorem: $\sqrt{3^2 + 4^2} = \sqrt{25} = 5\text{ N}$.
The resultant direction is diagonal between the two forces, not along either original axis, which corresponds to the direction OY shown in the diagram.
This matches the calculation and vector addition for forces at right angles.
Thus, the resultant force is $5\text{ N}$ along OY.
Question 2
| the initial acceleration of the ball | the final acceleration of the ball | |
|---|---|---|
| A | 0 | 0 |
| B | 0 | g |
| C | g | 0 |
| D | g | g |
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Initially, at the moment of release, the only force acting on the ball is its weight, resulting in an acceleration equal to the acceleration of free fall \( g \).
As the ball accelerates, air resistance increases until it balances the weight of the ball, at which point the resultant force becomes zero.
When the resultant force is zero, the acceleration is zero and the ball continues to move at a constant terminal velocity.
Therefore, the initial acceleration is \( g \) and the final acceleration is \( 0 \), which corresponds to row C.
Question 3

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The distance travelled is found from the area under the speed-time graph between 0 and 5 s.
The graph shows a horizontal line at a constant speed of $5\text{ m/s}$, forming a rectangle with time.
Calculate the area: distance $=$ speed $\times$ time $= 5\text{ m/s} \times 5\text{ s}$.
This gives a total distance moved of $25\text{ m}$.
Therefore, the correct answer is Option C.
Question 4
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Mass is a scalar quantity measuring the amount of matter in an object, not a force.
Weight is a vector quantity defined as the gravitational force acting on a mass (W = mg).
A balance compares gravitational forces to measure mass, meaning both masses and weights can effectively be compared using one.
Therefore, the statement “Mass is a force” is the incorrect option.
This matches Option B.
Question 5
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Weight is calculated using the formula $W = mg$. Since $g_{\text{Earth}} > g_{\text{Mars}}$, for both spheres to have the same weight ($10\text{ N}$), sphere Q on Mars must compensate for the lower field strength with a greater mass.
Therefore, the mass of sphere P is less than the mass of sphere Q.
Option B is the correct statement.
Question 6
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Density ($\rho$) is defined by the equation $\rho = \frac{m}{V}$, representing the mass ($m$) of a substance contained within a unit volume ($V$).
Therefore, calculating the density of any material strictly requires knowing its total mass and the volume it occupies.
Options involving area or weight are incorrect as area is not a measure of three-dimensional space and weight is a force dependent on gravity, not an inherent property like mass.
This matches Option B.
Question 7

| direction of acceleration of the truck | reason | |
|---|---|---|
| A | left | the driving force is greater than the resistive forces on the truck |
| B | left | the driving force is smaller than the resistive forces on the truck |
| C | right | the driving force is greater than the resistive forces on the truck |
| D | right | the driving force is smaller than the resistive forces on the truck |
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Before the rope breaks, the truck moves at constant speed, so the forward driving force is balanced by the total resistive forces plus the tension from towing the car.
When the rope breaks, the tension force is suddenly removed, making the driving force greater than the remaining resistive forces on the truck alone.
This creates a resultant force acting to the right, causing the truck to accelerate in the direction of the driving force.
Therefore, the initial acceleration is to the right because the driving force exceeds the resistive forces.
Question 8
A smaller mass is then added at the position shown.

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Adding the smaller mass on the right side creates an additional clockwise moment about the pivot.
To restore balance, the total clockwise moment must equal the total anticlockwise moment.
Moving mass X towards the pivot decreases its anticlockwise moment, allowing the beam to balance again.
This application of the principle of moments confirms that option B is correct.
Question 9
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Using Newton’s second law, calculate the acceleration: $a = \frac{F}{m} = \frac{2.0}{3.0} = \frac{2}{3} \text{ m/s}^2$.
The change in velocity is given by $\Delta v = a \times t = \frac{2}{3} \times 6.0 = 4.0 \text{ m/s}$.
This matches Option C.
Question 10

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
In a mobile phone, the chemical energy store of the battery is depleted to transfer energy to other components.
This transfer occurs via an electric current flowing through the phone’s circuits.
According to the syllabus, energy transferred by electrical currents is defined as electrical work done.
Therefore, the process linking the chemical store to the electrical components is correctly described as electrical work done.
This aligns with Option A.
Question 11
A wind turbine is 30% efficient and has an output of 2.5 MW of electrical power.
What is the power input to the turbine?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Efficiency is the ratio of useful power output to total power input: $\text{Efficiency} = \frac{\text{Output Power}}{\text{Input Power}}$.
Rearranging this gives $\text{Input Power} = \frac{\text{Output Power}}{\text{Efficiency}}$.
Substitute the given values: $\text{Input Power} = \frac{2.5\text{ MW}}{0.30} = 8.33\text{ MW}$.
Rounding to two significant figures (consistent with 2.5 MW) yields $8.3\text{ MW}$.
This corresponds to Option B.
Question 12
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Power is defined as the rate of doing work, given by the equation $P = \frac{W}{t}$ or $P = \frac{\Delta E}{t}$.
To calculate the power, the total work done ($W$) against gravity must be known along with the time ($t$) taken.
While work itself depends on force and height, power specifically requires the work done and the time taken only.
Therefore, the two essential quantities are work done and time taken, matching Option D.
Question 13
The density of sea water is \(1030 \, \text{kg/m}^3\). The gravitational field strength on the Earth is \(9.8 \, \text{N/kg}\). Atmospheric pressure is \(101 \, 000 \, \text{Pa}\).
At which depth in sea water is the total pressure due to the atmosphere and the water equal to \(513 \, 000 \, \text{Pa}\)?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Total pressure \(P_{\text{total}} = P_{\text{atm}} + \rho g h\). Rearranging gives \(h = \frac{P_{\text{total}} – P_{\text{atm}}}{\rho g}\).
Substitute the values: \(h = \frac{513000 – 101000}{1030 \times 9.8} = \frac{412000}{10094} \approx 40.8 \text{ m}\).
This is the depth at which the water pressure adds to atmospheric pressure to reach the total.
The calculation uses the supplement equation \(\Delta p = \rho g \Delta h\) from topic 1.8.
Question 14
Four students describe the phrase ‘absolute zero’ during a lesson on the particle model.
Which student is correct?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Absolute zero is defined as $0\text{ K}$ or $-273^{\circ}\text{C}$, the theoretical temperature at which particles have minimum internal energy and cease to move.
It is the lowest possible limit of the thermodynamic temperature scale and cannot be fully reached in practice.
Option A correctly identifies this fundamental concept from the kinetic particle model.
Other options incorrectly describe phase changes or unrelated energy forms.
Therefore, Option A is the accurate statement.
Question 15
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Thermal expansion occurs because particles vibrate more and move further apart when heated.
In solids, strong intermolecular forces restrict the particles, resulting in relatively small expansion.
In gases, the intermolecular forces are negligible, so particles can move much further apart, causing significantly greater expansion.
Option B correctly identifies that the weaker attractive forces in gases allow for a larger magnitude of expansion compared to solids.
This aligns with the kinetic particle model described in the syllabus.
Question 16
| the outside surface of the cup | the top of the cup | |
|---|---|---|
| A | black | covered with a lid |
| B | black | no lid |
| C | white | covered with a lid |
| D | white | no lid |
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• TOPIC 2.3.2: Convection (Explain convection in liquids and gases in terms of density changes)
▶️ Answer/Explanation
Detailed solution:
A white surface is a poor emitter of thermal radiation, so less heat is lost from the sides of the cup.
A lid prevents heat loss by convection currents and reduces evaporation from the liquid surface.
The combination of a reflective (white) outer surface and a sealed lid minimizes both radiation and convection.
This ensures the rate of thermal energy transfer to the surroundings is the lowest among the options.
Therefore, Cup C keeps the coffee warm for the longest time.
Question 17

| amplitude/cm | wavelength/cm | |
|---|---|---|
| A | 3 | 4 |
| B | 3 | 8 |
| C | 6 | 4 |
| D | 6 | 8 |
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Amplitude is the maximum displacement from the undisturbed position; here it is $3\text{ cm}$ (half the total vertical height of $6\text{ cm}$).
Wavelength is the distance between two adjacent crests or troughs; the diagram shows one full cycle covering $8\text{ cm}$.
Therefore, the amplitude is $3\text{ cm}$ and the wavelength is $8\text{ cm}$, corresponding to Option B.
Question 18
Optical fibres are used to transmit digital signals using infrared radiation. The average refractive index of the fibres is 1.50.
Which row describes a digital signal and gives the speed of infrared radiation in the fibres?
| digital signal | speed of infrared radiation in the fibres m/s | |
|---|---|---|
| A | a signal that consists of only two values | \( 2.0 \times 10^8 \) |
| B | a signal that consists of only two values | \( 4.5 \times 10^8 \) |
| C | a signal that consists of a continuous range of values | \( 2.0 \times 10^8 \) |
| D | a signal that consists of a continuous range of values | \( 4.5 \times 10^8 \) |
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
• TOPIC 3.3: Electromagnetic spectrum (Know the difference between a digital and analogue signal)
▶️ Answer/Explanation
Detailed solution:
A digital signal is characterized by having only two discrete values, typically representing 0 and 1.
The speed of light in a medium is calculated using \( v = \frac{c}{n} \), where \( c = 3.0 \times 10^8 \text{ m/s} \) and \( n = 1.50 \).
Substituting the values gives \( v = \frac{3.0 \times 10^8}{1.50} = 2.0 \times 10^8 \text{ m/s} \).
Therefore, the row with a two-value digital signal and a speed of \( 2.0 \times 10^8 \text{ m/s} \) is correct.
This matches Option A.
Question 19
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Monochromatic light is defined as light that consists of a single frequency or a single wavelength.
This is a specific term used in physics to distinguish pure spectral colours from white light, which is a mixture of all visible frequencies.
Options B, C, and D do not accurately describe this property; for instance, white light contains multiple frequencies and visible light spans a range of frequencies.
Therefore, Option A correctly matches the scientific definition of monochromatic light.
Question 20

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
When a beam of mixed light enters a glass prism, it undergoes dispersion due to different refractive indices for different wavelengths.
Blue light has a shorter wavelength and higher frequency than yellow light, so it experiences greater refraction and bends more towards the normal.
Therefore, the blue ray deviates further from the original path than the yellow ray, which matches the separation shown in diagram D.
This confirms Option D as the correct representation of dispersion through the prism.
Question 21
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Using the wave equation $\lambda = \frac{v}{f}$, substitute $v = 3.0 \times 10^8 \text{ m/s}$ and $f = 89 \times 10^6 \text{ Hz}$.
Calculate $\lambda = \frac{3.0 \times 10^8}{89 \times 10^6} = \frac{300}{89} \approx 3.37 \text{ m}$, which rounds to $3.4 \text{ m}$.
This wavelength is typical for radio waves in the VHF band, consistent with the electromagnetic spectrum.
Therefore, the wavelength of the 89 MHz signal is approximately $3.4 \text{ m}$.
Question 22
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The time of 2.2 s is for the sound to travel to the building and back.
The distance traveled one way is \( d = \frac{v \times t}{2} \).
Substituting the values: \( d = \frac{330 \times 2.2}{2} = 363 \text{ m} \).
Rounding to two significant figures gives approximately 360 m.
This corresponds to Option C.
Question 23

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Magnetic field lines always emerge from the north pole (N) and enter the south pole (S) of a magnet.
The lines must form continuous, closed loops and should never intersect each other.
In diagram A, the lines correctly curve outward from N and into S without crossing.
This matches the standard representation of the field around a bar magnet.
Question 24
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
When a plastic rod is rubbed with a dry cloth, friction causes negative charges (electrons) to transfer from the rod to the cloth.
The removal of these negatively charged electrons leaves the rod with a net positive charge.
Neutrons are located in the nucleus and are not transferred during simple friction.
Therefore, the rod becomes positively charged because it has lost electrons.
This corresponds to Option C.
Question 25
Which statement about electric current in a conductor is correct?
A. In a d.c. circuit, the electric current gradually decreases along the conductor.
B. In a d.c. circuit, the free electrons flow back and forth.
C. In an a.c. circuit, the electric current remains exactly the same all the time.
D. In an a.c. circuit, the flow of charge changes direction continually.
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Direct current (d.c.) flows in one direction only, while alternating current (a.c.) periodically reverses direction.
The current in a series d.c. circuit is the same at all points, not decreasing gradually.
Electrons in a.c. circuits oscillate back and forth rather than maintaining a steady flow in one direction.
Therefore, the statement that the flow of charge changes direction continually correctly describes a.c.
This matches the syllabus definition of a.c. in Topic 4.2.2.
Question 26
A piece of metal wire X with a uniform diameter has resistance \( R \).

A second piece of wire Y is made of the same metal and has a uniform diameter. Y has double the cross-sectional area of X and half the length of X.

What is the resistance of Y?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The resistance \( R \) of a wire is given by \( R = \rho \frac{L}{A} \).
For wire Y, the length \( L \) is halved (\( \frac{L}{2} \)) and the cross-sectional area \( A \) is doubled (\( 2A \)).
Substituting these values: \( R_Y = \rho \frac{L/2}{2A} = \rho \frac{L}{4A} \).
Since \( \rho \frac{L}{A} = R \), the new resistance is \( \frac{R}{4} \).
This matches Option A.
Question 27
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
First, calculate the current using \( P = IV \): \( I = \frac{1000\text{ W}}{240\text{ V}} = 4.17\text{ A} \).
Then, convert time to seconds: \( 1.0\text{ h} = 3600\text{ s} \).
Finally, find the charge using \( Q = It \): \( Q = 4.17\text{ A} \times 3600\text{ s} \approx 1.5 \times 10^4\text{ C} \).
This calculation directly applies the relationship between power, current, voltage, and charge.
Therefore, Option C is the correct answer.
Question 28

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
In the given parallel circuit, switch \( S_1 \) controls the main current from the power supply, so it must be closed to deliver power.
Switches \( S_2 \) and \( S_3 \) control the individual branches containing each lamp, so both must be closed to complete the path for each lamp.
Therefore, closing \( S_1 \), \( S_2 \), and \( S_3 \) ensures that both lamps are connected to the power supply and can light up.
This matches Option D.
Question 29

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The circuit contains a thermistor, which is a type of resistor whose resistance decreases as its temperature increases.
When the environmental temperature rises, the thermistor’s resistance falls, allowing more current to flow in the circuit.
This increased current activates the switching component (such as a relay or transistor), which then turns on the lamp.
Conversely, a decrease in temperature would increase the resistance and reduce the current, keeping the lamp off.
Options A and C relate to a light-dependent resistor (LDR), which is not the primary sensor shown here.
Question 30
A step-down transformer is 100% efficient. It has an input voltage of 240 V a.c. and an output voltage of 60 V a.c. The current in the primary coil is 0.50 A.
What is the current in the secondary coil?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
For a 100% efficient transformer, the power in the primary coil equals the power in the secondary coil, given by $V_p I_p = V_s I_s$.
Substitute the known values: $(240\text{ V}) \times (0.50\text{ A}) = (60\text{ V}) \times I_s$.
Calculate the primary power: $240 \times 0.50 = 120\text{ W}$.
Solve for the secondary current: $I_s = \frac{120\text{ W}}{60\text{ V}} = 2.0\text{ A}$.
This matches Option C.
Question 31

| direction of field | relative strength of field | |
|---|---|---|
| A | X | greater at P than Q |
| B | Y | greater at P than Q |
| C | X | greater at Q than P |
| D | Y | greater at Q than P |
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The magnetic field around a straight current-carrying wire forms concentric circles, and its direction (X or Y) is determined by the right-hand grip rule.
The strength of the magnetic field decreases as the distance from the wire increases.
Since point Q is closer to the wire than point P, the field strength is greater at Q than at P.
Based on the diagram and the application of the right-hand rule, the correct direction corresponds to Y.
Therefore, Option D (direction Y, greater at Q) is correct.
Question 32
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
An electromagnet uses a coil to create a magnetic field, while relays and loudspeakers operate based on the force exerted on a current-carrying conductor in a magnetic field.
A potential divider, however, consists of resistors arranged to provide a fraction of the input voltage and does not require any magnetic interaction to function.
Therefore, the magnetic effect of a current is not used in a potential divider.
Question 33
| total mass of fission products compared to original nuclide | total mass of fusion products compared to original nuclides | |
|---|---|---|
| A | same | same |
| B | more | less |
| C | less | more |
| D | less | less |
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
In both nuclear fission (splitting of a heavy nucleus) and nuclear fusion (joining of light nuclei), the total mass of the final products is less than the total mass of the original reactants.
This difference in mass, often called the mass defect, is converted into a large amount of energy as described by the equation \( E = mc^2 \).
This energy release is the principle behind the power of atomic bombs, nuclear reactors, and the Sun.
Therefore, for both processes, the mass of the products is less than the mass of the original components.
This corresponds to Option D.
Question 34
| nucleus | number of protons | number of neutrons |
|---|---|---|
| X | 3 | 3 |
| Y | 3 | 4 |
| Z | 4 | 3 |
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Isotopes are defined as nuclei of the same element that have an identical proton number ($Z$) but a different nucleon number ($A$) due to varying numbers of neutrons.
Nucleus X and Nucleus Y both contain $3$ protons, identifying them as the same element (Lithium), while differing in neutron count ($3$ vs $4$).
Nucleus Z contains $4$ protons, which corresponds to a different element (Beryllium) entirely.
Therefore, only X and Y represent isotopes of the same element.
This corresponds to Option B.
Question 35
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
In β⁻ emission, a neutron in the unstable nucleus decays into a proton, an electron, and an antineutrino.
The proton remains in the nucleus, increasing the atomic number by 1, while the electron is ejected as the beta particle.
This process reduces the neutron-to-proton ratio, moving the nucleus toward stability.
Therefore, the correct description is that a neutron transforms into a proton and an electron.
Question 36
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
First, find the initial count rate from the source by subtracting background radiation: $200 – 20 = 180$ counts/min.
After one half-life (8 days), the activity of the source halves, so the source count rate becomes $180 \div 2 = 90$ counts/min.
The question asks specifically for the count rate due to the source, not the total detected rate including background.
Therefore, the correct value is $90$ counts per minute, which corresponds to Option B.
Question 37
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Option A is incorrect: The four inner planets are rocky and small, while the four outer planets are gaseous and large.
Option B is incorrect: Most planets in the Solar System have moons (natural satellites) orbiting them.
Option C is correct: According to the syllabus, the Solar System contains minor planets including dwarf planets such as Pluto.
Option D is incorrect: The Solar System contains only one star, which is the Sun.
Therefore, the statement that Pluto is a dwarf planet is the only correct option.
Question 38
| shape of orbit | position of the Sun | |
|---|---|---|
| A | circular | centre of orbit |
| B | circular | not at centre of orbit |
| C | elliptical | centre of orbit |
| D | elliptical | not at centre of orbit |
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Comets travel in highly elongated, elliptical paths rather than perfect circles.
In an elliptical orbit, the Sun is located at one of the two foci of the ellipse, meaning it is offset from the geometric centre.
This positioning explains why comets move much faster when they are closer to the Sun (near perihelion) and slower when they are farther away.
Option D correctly identifies both the shape as elliptical and the Sun’s position as not at the centre.
Question 39
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Stars generate energy through the process of nuclear fusion, not fission.
In the core of a stable star like the Sun, hydrogen nuclei (protons) combine under extreme temperature and pressure to form helium nuclei.
This reaction results in a mass defect, which is converted into a large amount of energy according to $E = mc^2$.
Therefore, the correct description of the reaction is the nuclear fusion of hydrogen into helium, making Option D the accurate choice.
Question 40
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
The Hubble constant (\(H_0\)) is precisely defined as the constant of proportionality between the recessional velocity (\(v\)) of a galaxy and its distance (\(d\)) from the observer.
This relationship is given by the equation \(H_0 = \frac{v}{d}\) and is fundamental evidence for the expansion of the Universe.
Option A correctly states this ratio, which is distinct from the redshift value itself (Option B) or the reciprocal relationship used to estimate the age of the Universe (Option D).
Therefore, the definition required matches Option A.
