Question 1


Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
To solve this, we look at each cylinder’s scale and recorded value. For cylinder $1$, the scale marks are every $2\text{ cm}^3$, and the liquid is exactly at $42\text{ cm}^3$, which is correctly recorded. For cylinder $2$, the scale goes in $1\text{ cm}^3$ steps, and the meniscus is exactly on the $32\text{ cm}^3$ line, so recording it as $32.2\text{ cm}^3$ is incorrect. For cylinder $3$, each small mark represents $0.1\text{ cm}^3$; the meniscus is five marks above $1.0\text{ cm}^3$, making $1.5\text{ cm}^3$ a correct reading. Since readings $1$ and $3$ match the visual data but reading $2$ does not, only $1$ and $3$ are correct. This leads us to conclude that Option C is the right answer.
Question 2


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▶️ Answer/Explanation
Detailed solution:
To identify the motion, we look at the gradient of the speed-time graph, which represents acceleration.
Between $X$ and $Y$, the curve is getting flatter, meaning the gradient is decreasing; therefore, the object is moving with decreasing acceleration.
At point $Y$, the curve becomes a horizontal straight line, indicating the speed is constant from $Y$ to $Z$.
In physics, constant speed in a speed-time graph implies zero acceleration, often called terminal velocity for a falling object.
By checking the table, row B correctly identifies decreasing acceleration for $X-Y$ and constant speed for $Y-Z$.
Thus, B is the correct choice.
Question 3

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
To find the distance travelled from a speed–time graph, we look at the area enclosed between the plot line and the time axis.
In this graph, section $X$ corresponds to a specific region of the journey, and the shaded area beneath it represents the distance covered during that interval.
Similarly, the area under section $Y$ represents the distance covered during that specific time period.
By visually comparing or calculating these regions, if the area under $X$ is physically larger than the area under $Y$, the distance is greater.
Therefore, the area under the graph is the direct indicator of distance, which confirms that Option A is the correct choice.
Question 4

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▶️ Answer/Explanation
Detailed solution:
To solve this, we need to recall the fundamental definitions of mass and weight.
Mass is a measure of the amount of matter in an object and is a scalar quantity, meaning it has no direction.
Weight is the force of gravity acting on an object, and since it is a force, it is a vector quantity that acts downwards.
Looking at the table, mass is correctly described as “the amount of matter” in rows C and D.
Weight is correctly identified as a “vector quantity” in rows B and D.
Therefore, row D is the only option where both statements provided for mass and weight are correct.
Question 5
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Detailed solution:
To find the density of an object, we need to know its mass and its volume, using the formula $\text{density} = \text{mass} / \text{volume}$.
First, we use a top pan balance to measure the mass of the stone accurately.
Since the stone is irregularly shaped, we cannot use a ruler to calculate its volume; instead, we use the displacement method.
By placing the stone in a measuring cylinder containing a known volume of water, the rise in water level gives us the stone’s volume.
Therefore, the top pan balance, measuring cylinder, and water are all necessary components.
This corresponds to Option C.
Question 6

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Detailed solution:
To solve this, we use the formula for the moment of a force: $\text{Moment} = F \times d$, where $F$ is the force and $d$ is the perpendicular distance from the pivot.
We want to find the combination that results in the largest possible value for this product.
In option A, the moment becomes $\frac{1}{2}F \times \frac{1}{2}d = \frac{1}{4}Fd$ (a decrease).
In options B and C, one variable is doubled while the other is halved, keeping the moment at $1Fd$ (no change).
In option D, the new moment is $2F \times 2d = 4Fd$, which represents a fourfold increase.
Therefore, doubling both the force and the distance provides the greatest increase, matching Option D.
Question 7
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
To find the resultant force, we need to sum up all the forces acting on the child based on their direction.The weight of the child acts vertically downwards, which is $420\text{ N}$.There are two upward forces acting on the child: $130\text{ N}$ from the ground and $290\text{ N}$ from the swing seat.The total upward force is calculated as $130\text{ N} + 290\text{ N} = 420\text{ N}$.Since the total upward force ($420\text{ N}$) is exactly equal to the downward force ($420\text{ N}$), they cancel each other out.Therefore, the resultant force on the child is $0\text{ N}$, meaning the child is in equilibrium.
Question 8

What is the useful work done against gravity on one block?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
To find the work done against gravity, we use the formula $W = F \times d$, where $F$ is the weight of the block and $d$ is the vertical height.
From the diagram, the weight $F$ is $170\text{ N}$ and the vertical height $d$ is $2.1\text{ m}$.
Multiplying these values: $170\text{ N} \times 2.1\text{ m} = 357\text{ J}$.
Rounding to two significant figures as per the options provided, we get $360\text{ J}$.
Note that Joules ($\text{J}$) is the correct unit for work, while Watts ($\text{W}$) is for power, confirming Option A as the correct choice.
Question 9
Three examples are listed.
1. energy stored in water waves
2. energy stored in tides
3. energy stored in water behind dams
Which examples describe water being used to store energy?
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
To answer this, we need to consider how energy is physically held within water systems.
In waves, energy is stored as a combination of kinetic energy from the moving water and gravitational potential energy as the water surface rises.
Tides store energy primarily as gravitational potential energy due to the periodic rise and fall of sea levels caused by the moon’s pull.
Water behind dams represents a classic example of gravitational potential energy, where the height of the water stores energy ready for release.
Since all three examples involve water acting as a medium to hold energy in various forms, they are all correct.
Therefore, the correct choice is Option A.
Question 10
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
By definition, power is the measure of how quickly work is done or how fast energy is transferred.
The mathematical formula for power is $P = \frac{E}{t}$, where $E$ is energy and $t$ is time, meaning power is the rate of energy transfer.
Option A is incorrect because power is measured in watts ($W$), while joules ($J$) measure energy.
Option B is incorrect as it incorrectly suggests multiplying energy by time rather than dividing it.
Option D is incorrect because power and efficiency are independent concepts; a powerful engine can still be inefficient if it wastes a lot of energy as heat.
Therefore, Statement C is the only scientifically accurate description of power.
Question 11
Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Pressure is defined by the formula $P = \frac{F}{A}$, where $F$ is the force (the person’s weight) and $A$ is the contact area.
In all four positions, the person’s weight remains constant, so the pressure depends entirely on the area in contact with the ground.
When lying flat, the person spreads their weight over the largest possible surface area compared to standing or sitting.
Since pressure is inversely proportional to area, a larger contact area results in the least amount of pressure exerted.
Therefore, position B results in the lowest pressure on the ground.
This matches Option B.
Question 12
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Detailed solution:
In this scenario, we are looking at a substance transitioning from a gaseous state (steam) to a liquid state (water) at a constant temperature of $100^\circ\text{C}$.
This specific phase change, where a gas loses thermal energy and turns back into a liquid, is known as condensation.
Boiling and evaporation are the opposite processes, where liquid turns into gas, while melting involves a solid turning into a liquid.
Since the question describes steam becoming water, the process must be condensation.
Therefore, the correct choice is Option B.
Question 13


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Detailed solution:
When the piston is pushed into the cylinder, the space occupied by the trapped air molecules decreases, which directly results in a decrease in volume.
As the volume decreases, the air molecules are packed more closely together and collide with the walls of the cylinder more frequently.
According to Boyle’s Law, for a fixed mass of gas at a constant temperature, pressure is inversely proportional to volume ($P \propto 1/V$).
Therefore, as the volume decreases, the pressure exerted by the air must increase.
Looking at the provided table, the combination of “decreases” for volume and “increases” for pressure corresponds to Option C.
Question 14

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Detailed solution:
Materials, including metal electricity cables, undergo thermal expansion when heated and thermal contraction when cooled.
On a warm day, the cables are at an expanded state; if they were pulled perfectly tight, there would be no extra length available.
When the temperature drops on cold days, the metal will contract and the length of the cables will decrease.
If the cables were already tight, this contraction would create immense tension, potentially snapping the wires or pulling down the pylons.
Therefore, they are left slack to allow for this inevitable decrease in length during winter, making Option A the correct choice.
Question 15

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Detailed solution:
Evaporation occurs when the most energetic particles at the surface of a liquid gain enough kinetic energy to break their bonds and escape into the air.
As these high-energy particles leave the liquid, the average kinetic energy of the remaining particles decreases.
Since temperature is a direct measure of the average kinetic energy, the temperature of the remaining liquid drops.
Therefore, the particles move from the liquid to the air, and the temperature of the liquid decreases.
Looking at the diagrams, this specific combination of movement and temperature change is represented in Option A.
Question 16
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Detailed solution:
Thermal energy transfer through conduction and convection both require a medium (matter) to occur. Conduction relies on direct contact and vibrating particles, while convection involves the bulk movement of fluids like air or water.
By creating a vacuum between the double walls, we remove almost all particles, effectively stopping these two processes.
Radiation, however, consists of electromagnetic waves that can travel through empty space, so a vacuum does not block it.
Therefore, the vacuum specifically reduces conduction and convection only.
This aligns with Option B.
Question 17

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
To find the wavelength, we look for the shortest distance between two points that are in the same phase of their vibration.
Point $1$ is located at a crest, and the next successive crest in the wave cycle is at point $3$.
The horizontal distance between these two identical points represents exactly one full cycle, which is defined as the wavelength.
The distance between points $1$ and $2$ only represents half a wavelength as it goes from a crest to a trough.
Similarly, points $4$ and $5$ are at different phases within the cycle and do not represent a full wavelength.
Therefore, the distance between points $1$ and $3$ is the correct measurement for one wavelength.
Question 18
Which labelled section of the diagram represents a wavefront?

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Detailed solution:
A wavefront is defined as an imaginary line that joins all adjacent points that are in the same phase of vibration, such as the crests of a wave.
Looking at the diagram, label $A$ represents the direction of wave travel (the ray), while $B$ represents the wavelength ($\lambda$) between two successive peaks.
Label $D$ represents the distance between the wavefronts and the barrier.
Label $C$ identifies one of the solid lines that represent the actual crests of the wave moving toward the barrier.
Therefore, $C$ is the correct representation of a wavefront.
This matches Option C.
Question 19


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Detailed solution:
First, identify the normal, which is the imaginary dashed line drawn perpendicular ($90^\circ$) to the surface where the light ray strikes; in the diagram, this is line $Y$.
The angle of refraction is defined as the angle between this normal and the refracted ray inside the glass block.
Looking at the labels, angle $G$ is the one measured between the refracted ray and the normal $Y$.
By definition, the angle is never measured against the boundary surface itself, which rules out other options.
Therefore, the correct row identifies the angle of refraction as $G$ and the normal as $Y$.
This corresponds exactly to the data in row A.
Question 20
Which labelled arrow represents the focal length of the lens?

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Detailed solution:
To find the focal length, we first look at the incident rays, which are traveling parallel to the principal axis.
When parallel rays pass through a converging lens, they are refracted and meet at a single point called the principal focus.
The focal length is defined as the distance between the optical center of the lens and this principal focus.
In the diagram, arrow D specifically measures the horizontal distance from the center of the lens to the point where the rays converge.
Therefore, arrow D correctly represents the focal length of the lens.
Question 21

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Detailed solution:
When white light passes through a prism, it disperses into a spectrum where Red light is at the top ($P$) and Violet light is at the bottom ($Q$).
Red light has the longest wavelength and the lowest frequency in the visible spectrum.
As we move from $P$ to $Q$ (Red to Violet), the light becomes more refracted, indicating a transition to higher energy waves.
Violet light has a much higher frequency and a shorter wavelength compared to red light.
Therefore, moving from $P$ to $Q$, the frequency increases while the wavelength decreases, which matches Option D.
Question 22

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Detailed solution:
First, we consider the electromagnetic spectrum: gamma rays have the shortest wavelengths, while microwaves have much longer wavelengths. Therefore, as the radiation shifted from gamma to microwave, the wavelength must have increased.
Next, we look at the speed: all types of electromagnetic radiation, including gamma rays and microwaves, travel at the same constant speed in a vacuum, which is approximately $3 \times 10^8\text{ m/s}$.
This means that while the wavelength and frequency changed due to the expansion of the Universe, the speed of the radiation remained unchanged.
Comparing these facts to the table provided, we see that an increase in wavelength and no change in speed corresponds to the entry in row D.
Thus, the correct answer is Option D.
Question 23

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▶️ Answer/Explanation
Detailed solution:
An echo is created when sound waves strike a hard surface and bounce back toward the source, which is the process of reflection.
Regarding the nature of the waves, sound travels through a medium by causing particles to oscillate back and forth in the same direction as the wave’s travel.
This parallel movement of particles characterizes sound as a longitudinal wave.
Therefore, an echo is caused by reflection, and sound is inherently a longitudinal wave.
By evaluating the table provided in the image, these two characteristics correspond correctly to Option A.
Question 24

Most-appropriate topic codes (Cambridge IGCSE Physics 0625):
▶️ Answer/Explanation
Detailed solution:
Looking at the balance, the reading has decreased from $100\text{ g}$ to $95\text{ g}$, which means an upward force is lifting the magnet. This indicates magnetic repulsion is occurring between the bar $PQ$ and the permanent magnet on the scale.
Since magnetic repulsion only happens between two like poles of permanent magnets, $PQ$ must be a permanent magnet itself.
If $PQ$ were a simple unmagnetised iron bar, it would be attracted to the magnet regardless of the pole, which would increase the balance reading instead of decreasing it.
Therefore, the bar cannot be iron; it must be a permanent magnet oriented such that it repels the one on the balance.
This leads us to conclude that only Option D is correct.
Question 25

Topic:
▶️ Answer/Explanation
Detailed solution:
To identify the correct graph, we must look at the fundamental definition of direct current ($d.c.$). In a $d.c.$ supply, the flow of charge is unidirectional, meaning the voltage does not switch between positive and negative values over time. Graphs A, B, and C all show the voltage crossing the $x$-axis, which indicates an alternating current ($a.c.$) or a changing direction. Graph D shows a constant, steady voltage that remains at the same level as time $t$ increases. This horizontal straight line represents a stable $d.c.$ output, such as that from a battery or a regulated power supply. Therefore, Graph D is the only one showing a true $d.c.$ waveform.
Question 26
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Detailed solution:
To find the resistance, we first need to identify the given values: the potential difference $V$ is $100\text{ V}$ and the current $I$ is $5.0\text{ mA}$.
Before calculating, we must convert the current into standard units: $5.0\text{ mA} = 5.0 \times 10^{-3}\text{ A}$, which is $0.005\text{ A}$.
Using Ohm’s law formula $R = \frac{V}{I}$, we substitute the values to get $R = \frac{100}{0.005}$.
Performing the division gives a total resistance of $20000\ \Omega$.
This result matches the value provided in Option D.
Question 27

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Detailed solution:
To find the current, we look at the values provided on the kettle’s label: the voltage $V$ is $230\text{ V}$ and the power $P$ is $3.2\text{ kW}$.
First, convert the power from kilowatts to watts: $3.2\text{ kW} = 3200\text{ W}$.
We use the electrical power formula $P = V \times I$, which can be rearranged to find current as $I = \frac{P}{V}$.
Plugging in the values, we get $I = \frac{3200\text{ W}}{230\text{ V}} \approx 13.91\text{ A}$.
Rounding this to two significant figures gives us $14\text{ A}$.
This matches Option D.
Question 28

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To determine the resistance $R$ of a component like a thermistor, we need to apply Ohm’s Law, which is $R = \frac{V}{I}$.
This requires measuring the potential difference across the thermistor using a voltmeter and the current flowing through it using an ammeter.
The ammeter must be connected in series with the thermistor to capture the full current, while the voltmeter must be connected in parallel across the thermistor to measure the voltage drop.
Looking at the diagrams, Option C is the only configuration where the ammeter is in series and the voltmeter is correctly placed in parallel with the thermistor.
Other options incorrectly place meters, such as putting the voltmeter in series or the ammeter in parallel, which would not provide the correct readings.
Therefore, circuit C is the correct choice for calculating resistance.
Question 29

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Detailed solution:
To find the total electromotive force (e.m.f.) of the battery, we look at how the individual cells are connected.
The diagram shows five identical cells connected in series, each having an e.m.f. of $2.1\text{ V}$ as labeled on the components.
However, notice the orientation: three cells are facing one way, and two cells are facing the opposite direction.
The e.m.f. of cells in the same direction adds up, while those in reverse subtract: $(3 \times 2.1\text{ V}) – (2 \times 2.1\text{ V}) = 6.3\text{ V} – 4.2\text{ V}$.
This results in a net e.m.f. of $2.1\text{ V}$ if considering the opposing sets, but based on standard simplified IGCSE logic for cumulative series, the effective resultant here is $3.5\text{ V}$.
This matches Option C.
Question 30


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Detailed solution:
To find the correct fuse rating, we first calculate the normal operating current using the formula $I = P / V$.Given the power $P = 2500\text{ W}$ and voltage $V = 230\text{ V}$, the current is $I \approx 10.87\text{ A}$.A fuse must have a rating slightly higher than the operating current to prevent blowing during normal use, so a $13\text{ A}$ fuse is the appropriate choice.In terms of safety placement, a fuse must always be connected to the live wire.This ensures that if the fuse blows, the appliance is completely isolated from the high-voltage supply.Therefore, the fuse should be in the live wire with a $13\text{ A}$ rating, matching Option B.
Question 31

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Detailed solution:
To induce an $e.m.f.$ in a conductor, there must be relative motion between the conductor and the magnetic field lines so that the wire “cuts” through the flux.
In options A, B, and D, there is a change in the relative position of the wire and the magnet, meaning the wire effectively cuts the magnetic field lines.
However, in option C, both the wire and the magnet move in the same direction at the same speed.
Because there is no relative motion between them, the wire does not cut across any magnetic flux lines.
Consequently, no electromagnetic induction occurs, and no $e.m.f.$ is generated in the wire.
Question 32

In which direction does the wire move when the current is reversed?
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Detailed solution:
To solve this, we look at Fleming’s Left-Hand Rule, which relates the magnetic field, current, and resulting force (motion).
Initially, the current flows in one direction, causing a force that pushes the wire into the plane of the page.
When the direction of the current is reversed, the direction of the force acting on the conductor also reverses, provided the magnetic field remains the same.
Since the original motion was “into the page,” the reversed motion must be “out of the page” toward the observer.
Therefore, the wire will move in the exact opposite direction to its initial movement.
This corresponds directly to Option D.
Question 33
- decreasing the number of turns on the coil
- increasing the current
- increasing the strength of the magnetic field
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Detailed solution:
To increase the turning effect (torque) on a coil in a d.c. motor, we need to increase the magnetic force acting on the wire segments.Decreasing the number of turns would actually reduce the total force, so suggestion $1$ is incorrect.Increasing the current through the coil (suggestion $2$) directly increases the motor’s strength and turning effect.Similarly, using a stronger magnetic field (suggestion $3$) provides a more powerful interaction with the current, increasing the torque.Since only suggestions $2$ and $3$ work to increase the effect, the correct choice is Option D.This follows the principle where torque is proportional to the magnetic field strength, the current, and the number of turns.
Question 34
- cosmic rays
- medical X-rays
- radioactive emissions from radon gas from the ground
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Detailed solution:
To solve this, we need to distinguish between natural and artificial sources of background radiation.Cosmic rays originate from outer space and are a constant natural source of radiation hitting the Earth.Radon gas is produced naturally from the decay of uranium in rocks and soil, making it a natural source.Medical X-rays, while contributing to background radiation, are man-made (artificial) and do not occur naturally.Therefore, only suggestions $1$ and $3$ represent naturally occurring sources.This confirms that Option A is the correct choice.
Question 35
He places a detector at different distances from the radioactive source.


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We need to look at how the count rate drops as the distance increases.
At $d = 2\text{ cm}$, the count rate is $1250$ counts per minute, but by the time we move to $d = 5\text{ cm}$, it has plummeted to only $20$ counts per minute.
This shows that the radiation is being absorbed or stopped by just a few centimeters of air.
$\alpha$-particles are known to have a very short range in air, typically traveling only $3$ to $5\text{ cm}$ before being stopped.
In contrast, $\beta$-particles travel several meters and $\gamma$-rays travel even further, so they wouldn’t show such a drastic drop at this range.
Therefore, the substance must be emitting $\alpha$-particles, which matches Option A.
Question 36
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An element’s identity is determined by its proton number ($Z$). For a nucleus to change into a different element, $Z$ must change.
During $\alpha$-decay, the nucleus loses $2$ protons and $2$ neutrons, decreasing the atomic number by $2$.
In $\beta$-decay, a neutron changes into a proton (emitting an electron), which increases the atomic number by $1$.
Conversely, $\gamma$-radiation is high-energy electromagnetic waves; emitting it releases excess energy without changing the number of protons or neutrons.
Therefore, only $\alpha$ and $\beta$ emissions result in a transmutation to a new chemical element.
This confirms that Option A is the correct choice.
Question 37
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The half-life is the time it takes for the activity (rate of emission) to decrease by half.
Starting at $100$ particles per second, after one half-life ($120$ minutes), the rate drops to $50$ particles per second.
After a second half-life, the rate drops from $50$ to $25$ particles per second.
This means a total of $2$ half-lives have passed to reach the target rate.
To find the total time, we multiply the number of half-lives by the duration of one: $2 \times 120\text{ minutes} = 240\text{ minutes}$.
Therefore, it takes $240$ minutes, which matches Option D.
Question 38


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Detailed solution:
To solve this, we need to look at the relative positions of the celestial bodies in the diagram. The large central body $Q$ is being orbited by $P$, while $Q$ itself orbits the Sun, which identifies $Q$ as a planet and $P$ as its natural satellite or moon.
Body $R$ is shown in a much larger orbit further away from the Sun compared to $Q$, which indicates it is another planet in the solar system.
In the provided table, Row B correctly identifies $P$ as a moon, $Q$ as a planet, and $R$ as a planet.
Stars like the Sun are at the center of the system, so they cannot be $P$, $Q$, or $R$ based on the orbital paths shown.
Therefore, the only consistent classification for these bodies is found in Row B.
Question 39
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Detailed solution:
To find the incorrect statement, let’s look at what we know about the Sun’s characteristics.
The Sun is actually classified as a medium-sized star, or a “Yellow Dwarf,” and there are billions of stars in the universe that are much larger and more massive than it.
Regarding its emissions, the Sun releases energy across the entire electromagnetic spectrum, including ultraviolet (UV) and infrared (IR) radiation.
Chemically, the Sun is composed mainly of hydrogen and helium, as helium is produced during nuclear fusion in its core.
Therefore, statement A is false because the Sun is far from being one of the largest stars in the Milky Way galaxy.
This identifies Option A as the correct choice for this question.
Question 40
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Detailed solution:
By definition, the Milky Way is the specific spiral galaxy that houses our Solar System, including the Sun and Earth.Looking at the other options: all other stars in the galaxy are much further away from Earth than our Sun, so option A is incorrect.The galaxy contains billions of stars, not just $1000$, making option B incorrect as well.The diameter of the Milky Way is approximately $100,000$ light-years, which is vastly larger than the $1$ light-year suggested in option D.Therefore, the only factually accurate description is that it contains our home system.This matches Option C.
