Physics 5054/12 — May/June 2025

Walkthrough

A scalar quantity has only magnitude (size). A vector quantity has both magnitude and direction.

Look at each list:

• A: mass is a scalar, distance is a scalar, weight is a vector. Because mass and distance are scalars, the list is not only vectors.
• B: momentum is mass × velocity, so it has the direction of the velocity; gravitational field strength is force per unit mass and points towards the centre of the Earth; displacement is distance in a stated direction. All three are vectors, so this is the correct list.
• C: energy is a scalar, force is a vector, velocity is a vector. Because energy is a scalar, the list is not only vectors.
• D: acceleration is a vector, but time and speed are scalars. Because time and speed are scalars, the list is not only vectors.

Key Takeaways

• Vectors have both magnitude and direction: displacement, velocity, acceleration, force, weight, momentum and gravitational field strength.
• Scalars have only magnitude: mass, distance, speed, time, energy, temperature and density.
• In an O Level multiple-choice question, check every quantity in the list, not just the first one.

Common Mistakes

• Thinking that distance is a vector because it sounds similar to displacement. Distance has no direction, so it is a scalar.
• Thinking that speed is a vector because it is related to velocity. Speed is the magnitude of velocity and has no direction.
• Confusing weight with mass. Weight is a force and is a vector; mass is a scalar.

Things to Be Careful About

• The word “only” in the question means every quantity in the chosen list must be a vector.
• Gravitational field strength is a vector even though it is often written as a number such as 10 N/kg; its direction is towards the centre of the Earth.

Walkthrough

The question provides a distance-time graph and asks which physical scenario matches it. The key to reading a distance-time graph is that its gradient at any point equals the speed of the object at that time.

1. Initial section: The curve starts at the origin $(0,0)$ with a steep positive gradient. A steep gradient means a high speed. So the object is moving quickly at the start.
2. Middle section: The curve bends so that the gradient becomes less steep over time. A decreasing gradient means the speed is decreasing. The object is slowing down (decelerating).
3. Final section: The curve becomes a horizontal line. The gradient of a horizontal line is zero. A speed of zero means the object has stopped. The distance is no longer changing, so it remains constant.

Now we match this behaviour to the options:

• Option A (ball thrown up and falling back): If distance is measured from the thrower, the distance would increase as the ball rises and then decrease as it falls back down. The graph would go up and then come back down to the time axis. This does not match.
• Option B (car speeding up): If the car is speeding up, its speed is increasing. The gradient of the distance-time graph would increase over time, producing a curve that bends upwards (concave up). This does not match.
• Option C (rock dropped from a cliff): A falling rock accelerates due to gravity, so its speed increases. Like option B, the gradient would increase, producing a curve that bends upwards. This does not match.
• Option D (train braking to a halt): A braking train has a decreasing speed, so the gradient of the distance-time graph decreases over time, producing a curve that bends downwards (concave down). When the train halts, its speed becomes zero, and the graph becomes horizontal. This perfectly matches the given graph.

Key Takeaways

• The gradient of a distance-time graph represents speed.
• An increasing gradient (curve bending upwards) means the object is accelerating.
• A decreasing gradient (curve bending downwards) means the object is decelerating.
• A horizontal line on a distance-time graph means the object is stationary (speed is zero).

Common Mistakes

• Confusing a distance-time graph with a speed-time graph. On a speed-time graph, a horizontal line means constant speed, not a stopped object. Here, the y-axis is distance, so a horizontal line means distance is not changing, i.e., the object has stopped.
• Thinking a curved line always means constant acceleration. The curve here shows decreasing acceleration (deceleration), not constant acceleration. Constant acceleration would be a straight line on a velocity-time graph, or a parabola on a distance-time graph, but the key is the direction the curve bends.

Things to Be Careful About

• Always check the axis labels. The y-axis is distance, not speed. If it were speed, a horizontal line at a non-zero value would mean constant speed, not stopping.
• Remember that distance is a scalar and only increases or stays constant for a moving object in one direction. If an object changes direction (like the ball in option A), the distance-time graph would either show a decrease (if measuring displacement from a point) or a continuous increase (if measuring total path length), but never a horizontal line unless it stops permanently.

Walkthrough

The question asks which part of the speed–time graph shows the skydiver falling with terminal velocity.

1. Understand terminal velocity: As a skydiver falls, gravity pulls them down (weight) and air resistance pushes up (drag). Initially, weight is greater than drag, so the skydiver accelerates. As speed increases, drag increases. When drag equals weight, the resultant force is zero. By Newton's first law, the skydiver no longer accelerates and continues to fall at a constant speed. This constant speed is called terminal velocity.

2. Read the speed–time graph: The vertical axis is speed and the horizontal axis is time.

   • A sloping line means the speed is changing (acceleration or deceleration).
   • A horizontal line means the speed is not changing (constant speed, zero acceleration).

3. Match to the sections:

   • Section A: The curve rises steeply and then flattens. Speed is increasing, but the rate of increase (acceleration) is decreasing because drag is building up.
   • Section B: Speed is still increasing slightly, approaching a first terminal velocity (before the parachute opens).
   • Section C: The curve drops steeply. The parachute opens, greatly increasing air resistance. Drag is now much larger than weight, so the skydiver decelerates rapidly.
   • Section D: The line is horizontal. Speed is constant at a new, lower terminal velocity.

Since terminal velocity is constant speed, the correct section is D.

Key Takeaways

• Terminal velocity is constant speed, achieved when driving force (weight) equals resistive force (drag).
• On a speed–time graph, constant speed is always a horizontal line.
• Acceleration is the gradient of a speed–time graph; zero gradient means zero acceleration.

Common Mistakes

• Confusing speed–time with distance–time: On a distance–time graph, a horizontal line means the object is stationary. On a speed–time graph, a horizontal line means constant speed (not necessarily zero speed, unless it is on the time axis).
• Choosing A or B: These sections show the skydiver still accelerating (speed is increasing), so they are not terminal velocity.
• Choosing C: Section C shows rapid deceleration (negative acceleration) as the parachute opens, not constant speed.

Things to Be Careful About

• Always check the axis labels. If the vertical axis is 'speed' or 'velocity', a horizontal line is constant speed. If it is 'distance' or 'displacement', a horizontal line is stationary.
• Terminal velocity is not zero speed; it is the maximum constant speed reached during the fall. Section D is at a low but non-zero speed.
• The gradient of the graph gives acceleration. A horizontal line has a gradient of zero, confirming zero acceleration and thus constant speed.

Walkthrough

The question provides a speed-time graph with two shaded areas: X and Y. The graph shows a horizontal section (constant speed) followed by a straight line sloping down to zero (constant deceleration).

1. Area under the graph: In a speed-time graph, the area between the line and the time axis represents the distance travelled.
2. Area X: This rectangular area corresponds to the time interval before the driver applies the brakes. During this time, the car travels at a constant speed. The distance covered in this phase is called the thinking distance (the distance travelled during the driver's reaction time). Thus, student 1 is correct.
3. Area Y: This triangular area corresponds to the time interval while the brakes are applied and the car decelerates to rest. The distance covered in this phase is called the braking distance. Thus, student 2 is incorrect, as area Y is only the braking distance, not the total stopping distance.
4. Total stopping distance: The stopping distance is defined as the total distance travelled from the moment the driver sees the hazard until the car stops completely. This is the sum of the thinking distance and the braking distance. Therefore, Area X + Area Y represents the stopping distance, making student 3 correct.

Since students 1 and 3 are correct, the correct option is B.

Key Takeaways

• The area under a speed-time graph always represents the distance travelled.
• Stopping distance = thinking distance + braking distance.
• Thinking distance is the distance travelled at constant speed during the driver's reaction time (rectangular area on a speed-time graph).
• Braking distance is the distance travelled while the car is decelerating to rest (triangular area on a speed-time graph for uniform deceleration).

Common Mistakes

• Confusing braking distance with stopping distance. Students often think the deceleration area alone is the stopping distance.
• Misinterpreting the graph axes. Remember that speed-time graphs give distance from the area, not from the gradient (which gives acceleration).
• Forgetting that thinking distance occurs at constant speed, so it forms a rectangle, not a triangle.

Things to Be Careful About

• Ensure you read the graph axes correctly: speed-time graphs use area for distance, whereas distance-time graphs use gradient for speed.
• The term 'stopping distance' specifically includes both the thinking phase and the braking phase. 'Braking distance' is only the second phase.
• In this question, the deceleration is uniform (straight line), so the braking distance is a triangle. If deceleration were not uniform, the area would still be the braking distance, but it would be a curved shape.

Walkthrough

The question asks to identify the limit of proportionality on a load-extension graph. The graph plots load on the vertical axis against extension on the horizontal axis. Starting from the origin (0,0), the curve is a straight line up to point A. A straight line on a load-extension graph means that load is directly proportional to extension ($L \propto e$), which is the statement of Hooke's law. The limit of proportionality is defined as the point up to which this direct proportionality holds; it is the exact point where the straight line ends and the curve begins. Therefore, point A is the limit of proportionality. Beyond point A, the material deforms plastically. Point B represents the maximum load the material can support before necking begins, and point D is where the material fractures.

Key Takeaways

• On a load-extension graph, the initial straight-line region from the origin is where Hooke's law is obeyed.
• The limit of proportionality is the point where this straight line ends and the graph starts to curve.
• Beyond the limit of proportionality, the material undergoes plastic deformation and will not return to its original shape when the load is removed.

Common Mistakes

• Selecting point B: Point B is the maximum load (or yield point in some simplified diagrams). Students often confuse the limit of proportionality with the yield point or the point of maximum load.
• Selecting point D: Point D is the fracture point, where the material breaks, not the limit of proportionality.

Things to Be Careful About

• Ensure you are looking at a load-extension graph (or stress-strain, which has the same shape and features). The limit of proportionality is always the end of the initial straight-line portion from the origin.
• Do not confuse the limit of proportionality with the elastic limit; they are very close together on this graph, but the strict definition of the limit of proportionality is the end of the linear region where Hooke's law is obeyed.

Walkthrough

The question tells us there is no atmosphere on the Moon. This is the key clue: it means there is no air resistance or drag acting on the spheres. The only force on each sphere is its weight, which pulls it downwards.

For any object in free fall, the acceleration is the gravitational field strength $g$. Even though the heavier sphere has a larger weight, it also has a larger mass, so the acceleration $a = F/m = g$ is the same. This is why all objects fall with the same acceleration in a vacuum.

Both spheres are at rest at the same height and are dropped at the same time. With the same acceleration and the same starting conditions, they follow exactly the same motion and reach the ground at the same time. The fact that the spheres have identical volume is irrelevant; volume does not affect free fall.

Option D is correct. Options A and C are wrong because they suggest the acceleration depends on mass. Option B is wrong because it suggests the heavier sphere falls faster, which is the common misconception that only applies when air resistance is significant.

Key Takeaways

• In the absence of air resistance, all objects fall with the same acceleration regardless of their mass.
• The acceleration of free fall is the gravitational field strength $g$ of the planet or moon.
• A larger mass gives a larger weight, but the acceleration is still $g$ because the larger mass also needs a larger force to accelerate it.
• Air resistance is the reason objects of different mass or shape can fall at different speeds on Earth.

Common Mistakes

• Choosing B: thinking the heavier sphere hits the ground first. This is only true when air resistance is significant, but the question states there is no atmosphere.
• Choosing A or C: thinking acceleration depends on mass. In free fall, the acceleration is independent of mass.
• Ignoring the phrase 'no atmosphere' and applying Earth-like air resistance reasoning.

Things to Be Careful About

• The value of $g$ on the Moon is about 1.6 N/kg, but you do not need to know it; the same $g$ applies to both spheres.
• Do not confuse mass with weight. The heavier sphere has a larger weight, but its acceleration is still $g$.
• The identical volume is a distractor; it does not affect the motion in free fall.
• Both spheres start from rest at the same height, so the comparison is fair: same initial velocity, same height, same acceleration.

Walkthrough

The rock is submerged in water, so the increase in the measuring cylinder reading gives the rock's volume:

Density is mass per unit volume, but the question gives the rock's weight, not its mass. Use $W = mg$ with $g = 10\ \text{N/kg}$:

Then

This rounds to $2.6\ \text{g/cm}^3$, which is option D.

Options A and C give density in $\text{N/cm}^3$; density is mass per unit volume, not weight per unit volume, so those cannot be correct. Option B would come from an incorrect mass or volume.

Key Takeaways

• Density is defined as mass divided by volume.
• The volume of an irregular solid can be found by water displacement: the rise in the measuring cylinder reading equals the volume of the object.
• Weight and mass are different: $W = mg$, so mass is found by dividing weight by $g$.
• Units matter: here the answer is in $\text{g/cm}^3$ because the mass was converted to grams and the volume is in cubic centimetres.

Common Mistakes

• Using the weight $1.2\ \text{N}$ directly as the mass. This gives a wrong density and the wrong unit.
• Using the final reading $147\ \text{cm}^3$ as the rock's volume instead of the increase $47\ \text{cm}^3$.
• Choosing an answer in $\text{N/cm}^3$: density is not weight per unit volume.
• Forgetting to convert $0.12\ \text{kg}$ to $120\ \text{g}$ before dividing by $\text{cm}^3$.

Things to Be Careful About

• The displaced volume is the difference between the two readings, not the final reading.
• Use $g = 10\ \text{N/kg}$ as given in O Level questions unless another value is stated.
• Keep the units consistent: mass in grams with volume in $\text{cm}^3$ gives density in $\text{g/cm}^3$.
• The answer is quoted to 2 significant figures because the data has 2 significant figures.

Walkthrough

1. Identify the knowns: The problem gives us the mass $m = 20\ \text{kg}$ and the speed $v = 6.0\ \text{m/s}$ of the rock.
2. Recall the formula: The kinetic energy ($E_k$) of a moving object is given by the equation $E_k = \frac{1}{2}mv^2$.
3. Substitute and calculate: Substitute the known values into the equation:
   $E_k = \frac{1}{2} \times 20 \times (6.0)^2 = 10 \times 36 = 360\ \text{J}$.
4. Select the answer: The kinetic energy is $360\ \text{J}$, which is option C.

Key Takeaways

• The kinetic energy of an object is the energy it has due to its motion.
• The formula $E_k = \frac{1}{2}mv^2$ is fundamental and must be memorized.
• Ensure all quantities are in SI units (kg, m/s) before substituting into the formula.

Common Mistakes

• Forgetting to square the velocity: A common mistake is to calculate $\frac{1}{2} \times 20 \times 6.0 = 60\ \text{J}$, which is option A. Remember to square the velocity first.
• Using the wrong formula: Confusing this with momentum ($p=mv$) or gravitational potential energy ($E_p=mgh$).

Things to Be Careful About

• Always square the velocity ($v^2$) before multiplying by the mass and the half.
• The unit of energy is the joule (J).

Walkthrough

The question asks which row correctly pairs an advantage and a disadvantage of using nuclear fuel to generate electrical power.

Advantage: Nuclear fission releases energy without burning fuel, so no carbon dioxide is produced. This is a real environmental advantage compared with fossil fuels.

Disadvantage: The fission process leaves behind radioactive waste products that remain dangerous for a very long time and must be stored safely. This is the main drawback of nuclear power.

Now test the other statements:

• Renewable? No. Nuclear fuel (uranium) is a finite resource dug out of the ground, so it is non-renewable. Rows C and D are wrong on the advantage.
• Only works in some weather conditions? No. Nuclear reactors run continuously regardless of sunshine, wind or rain. That statement describes solar or wind power, not nuclear. Rows B and D are wrong on the disadvantage.

Only row A has both statements correct.

Key Takeaways

• Nuclear fuel is non-renewable but produces no carbon dioxide during generation.
• Its main disadvantage is the radioactive waste it produces.
• Weather-dependent generation ("only works in some weather conditions") applies to solar and wind power, not nuclear.

Common Mistakes

• Choosing C or D because a student thinks nuclear is renewable — it is not; uranium is a finite mined resource.
• Choosing B or D because a student confuses nuclear with solar or wind power, which do depend on weather.

Things to Be Careful About

• Read each statement in the row independently: check the advantage and the disadvantage separately before selecting the row.
• Remember that "does not produce carbon dioxide" is true for nuclear, but also for renewables — the distinguishing feature of nuclear is the radioactive waste.

Walkthrough

The question asks for the correct definition and unit of work done. Work is done when a force moves an object through a distance in the direction of the force. The amount of work done is calculated as force × distance moved in the direction of the force:

Its unit is the joule ($\text{J}$). One joule is the work done when a force of one newton moves an object one metre in the direction of the force, so $\text{J}$ is equivalent to $\text{N m}$, not $\text{N / m}$.

The phrase “energy transferred per unit time” is the definition of power, not work. Therefore only row C has both the correct definition and the correct unit.

Key Takeaways

• Work done = force × distance moved in the direction of the force.
• The unit of work done is the joule ($\text{J}$).
• Power is the rate of energy transfer, measured in watts ($\text{W}$), and is not the same as work done.
• $\text{N / m}$ is not a unit of work; $\text{N m}$ (newton metre) is equivalent to the joule.

Common Mistakes

• Choosing A or B: confusing work done with power. “Energy transferred per unit time” is power.
• Choosing D: knowing the correct definition but accepting $\text{N / m}$ as the unit. The unit of work is the joule ($\text{J}$), equivalent to $\text{N m}$, not $\text{N / m}$.
• Writing $\text{N / m}$ as the unit of work; work is force × distance, so the unit is $\text{N m} = \text{J}$.

Things to Be Careful About

• Read the whole row: both the definition and the unit must be correct.
• Remember the difference between $\text{N m}$ and $\text{N / m}$. $\text{N / m}$ would be a unit for something like force per distance, not work.
• The joule is the SI unit of work and energy; the watt is the unit of power.

Walkthrough

1. Forces ($F_\text{H}$ vs $F_\text{W}$):
   The problem states the weight of the nail is negligible. In physics problems of this type at O Level, we treat the nail as a light object transmitting force. According to Newton's laws, if an object has negligible mass, the resultant force on it is negligible ($F_{net} = ma \approx 0$). The force pushing the nail forward ($F_\text{H}$ from the hammer) must be balanced by the resistive force from the wood ($F_\text{W}$, effectively the force the nail exerts on the wood, by Newton's third law pairs acting on the nail). Thus, the force is transmitted directly: $F_\text{W} = F_\text{H}$. This eliminates options C and D.

2. Pressures ($p_\text{H}$ vs $p_\text{W}$):
   The formula for pressure is $p = \frac{F}{A}$, where $F$ is the force and $A$ is the area of contact.

   • $p_\text{H}$ is the pressure at the hammer-nail interface. The area here is the area of the hammer face or the nail head, which is relatively large.
   • $p_\text{W}$ is the pressure at the nail-wood interface. The area here is the cross-sectional area of the sharp tip of the nail, which is very small.
     Since $F_\text{H} = F_\text{W}$ (forces are equal) and $A_\text{W} < A_\text{H}$ (area at wood is smaller), the pressure $p_\text{W}$ is greater than $p_\text{H}$ ($p_\text{W} > p_\text{H}$).

3. Conclusion:
   Combining these, $F_\text{W} = F_\text{H}$ and $p_\text{W} > p_\text{H}$. This matches row A in the table.

Key Takeaways

• Force Transmission: In a system of light objects (negligible mass), the force applied to one end is transmitted to the other end. The force does not increase as it goes deeper; it stays the same (assuming no acceleration of the object itself).
• Pressure and Area: Pressure is inversely proportional to area for a given force. A sharp object (small area) exerts a large pressure, which is why nails and pins can penetrate surfaces.

Common Mistakes

• Thinking force increases: Candidates often choose C or D, believing that the force 'builds up' or is 'concentrated' as it goes into the wood. Force is not concentrated; it is transmitted. Only the pressure changes due to the area.
• Confusing force and pressure: Forgetting that pressure depends on area ($p=F/A$). If a candidate thinks the forces are different, they might get the pressure relationship wrong too.
• Unit errors: Not relevant here as it's a comparison, but always remember $p = F/A$.

Things to Be Careful About

• Negligible mass: The phrase 'weight of the nail is negligible' is the key to assuming $F_\text{W} = F_\text{H}$. If the nail had significant mass and was accelerating downwards rapidly, $F_\text{H} - F_\text{W} = ma$, so $F_\text{H} > F_\text{W}$. But for O Level, 'negligible' implies we treat them as equal.
• Area identification: Ensure you identify the correct contact areas. Hammer hits the top (large area), nail hits the bottom (small area).
• Vector vs Scalar: Force is a vector, pressure is a scalar (though it acts in a direction normal to the surface). The comparison is of magnitudes.

Walkthrough

The pressure in a liquid at a point depends on the weight of the liquid above it. The deeper you go, the more liquid is above you, so the pressure is greater. Similarly, if the liquid is denser, the liquid above you weighs more, so the pressure is greater. The equation is $p = h\rho g$. To increase pressure, both depth and density must increase, which is option D.

Key Takeaways

• Liquid pressure increases with depth.
• Liquid pressure increases with density.
• The equation $p = h\rho g$ summarises both relationships.

Common Mistakes

• Choosing an option where one quantity decreases: any decrease in depth or density lowers the pressure.
• Confusing pressure with force: pressure depends on the weight of liquid above, so density matters.

Things to Be Careful About

• Read the table carefully: the question asks which changes both increase the pressure, so both columns must be "increase".
• Remember $g$ is constant for the same liquid location, so only depth and density vary.

Walkthrough

1. Analyze option B (average speed): Heating the air transfers thermal energy to the gas particles, increasing their average kinetic energy. Since kinetic energy is proportional to the square of speed, the average speed of the particles increases. Option B is incorrect because it increases, not decreases.

2. Analyze option A (average force per collision): Force is the rate of change of momentum. Faster-moving particles have greater momentum, so when they bounce off the piston, the change in momentum per collision is larger. This means the average force exerted by each individual collision increases. Option A is incorrect because it increases.

3. Analyze option C (mass of air): The cylinder is sealed by the piston, meaning no air can enter or escape. By conservation of mass, the total mass of air inside the cylinder remains constant. Option C is incorrect because it stays the same.

4. Analyze option D (number of collisions per second): The piston is free to move and atmospheric pressure acts outside, so the air pressure inside remains constant (equal to the atmospheric pressure). Macroscopically, pressure is the total force per unit area. Microscopically, the total force on the piston is the product of the number of collisions per second and the average force per collision. Since the average force per collision has increased (from step 2) and the total pressure must remain constant, the number of collisions per second must decrease. Option D is correct.

Key Takeaways

• Temperature is a measure of the average kinetic energy of particles; an increase in temperature means an increase in average particle speed.
• Pressure in a gas is the result of particle collisions with the container walls. Total pressure depends on both the force per collision and the number of collisions per unit area per second.
• When a gas is heated at constant pressure (a movable piston), the volume increases. The particles move faster (increasing force per collision), but they are more spread out (decreasing collision frequency) so that the overall pressure remains unchanged.

Common Mistakes

• Confusing speed with collision frequency: A student might think that because particles move faster, they hit the piston more often. This ignores the fact that the volume has increased, reducing the number density of particles.
• Confusing mass with density: The density of the air decreases as the volume increases, but the total mass remains constant. Selecting C shows a misunderstanding of conservation of mass in a closed system.
• Ignoring the constant pressure condition: Failing to recognize that a free-moving piston against atmospheric pressure implies constant pressure is the key to solving the question.

Things to Be Careful About

• Always distinguish between the force per collision (which depends on particle speed) and the total force / pressure (which depends on both speed and collision frequency).
• Read the setup carefully: "free to move" and "atmospheric pressure acts outside" together mean the internal pressure is constant. If the piston were fixed, the pressure would increase, and the number of collisions per second would actually increase.
• In multiple-choice questions, systematically eliminating the clearly wrong options (B and C) makes the correct choice much easier to justify.

Walkthrough

The question gives a fixed mass of air at constant temperature and asks what happens to the pressure when the volume is reduced from $150\ \text{cm}^3$ to $100\ \text{cm}^3$. Both pressures are in kPa and both volumes in cm³, so no unit conversion is needed — the units cancel.

At constant temperature, the kinetic particle model tells us the air particles on average have the same kinetic energy, so they hit the walls with the same average force per collision. But if the volume is reduced, the same number of particles has less space, so the collision rate with the walls rises, and the pressure increases.

This is Boyle's law: for a fixed mass of gas at constant temperature, $p_1V_1 = p_2V_2$. Rearranging gives $p_2 = \frac{p_1V_1}{V_2} = \frac{120 \times 150}{100} = 180\ \text{kPa}$.

The volume decreased to two-thirds of the original, so the pressure must rise to three-halves of the original — an increase from 120 kPa to 180 kPa. This inverse relationship is the whole point: smaller volume means greater pressure.

The distractors:

• A (80 kPa) — the inverse ratio $\frac{120 \times 100}{150}$, i.e. treating pressure as proportional to volume instead of inversely proportional. The most common misconception.
• C (80 000 kPa) — the same inverse-ratio error as A but also with a power-of-ten slip.
• D (180 000 kPa) — the correct number 180 but expressed as if it were pascals rather than kilopascals ($180\ \text{kPa} = 180\,000\ \text{Pa}$), a unit error.

So B is the only option showing both the correct inverse ratio and the correct units.

Key Takeaways

• For a fixed mass of gas at constant temperature, pressure and volume are inversely proportional: halving the volume doubles the pressure.
• The working form is $p_1V_1 = p_2V_2$, and it works with any consistent pair of units.
• Identify a correct-ratio-with-unit-error distractor: $180\ \text{kPa}$ is NOT $180\,000\ \text{kPa}$ unless the question specifically asks for pascals.

Common Mistakes

• Using the ratio the wrong way round: $\frac{p_1V_2}{V_1}$ instead of $\frac{p_1V_1}{V_2}$, giving 80 kPa instead of 180 kPa.
• Confusing inverse proportion with direct proportion — as volume drops, pressure rises.
• Dropping the kilo prefix and writing 180 000 kPa instead of 180 kPa, or mixing kPa with Pa mid-calculation.

Things to Be Careful About

• Always ask yourself: should the answer be larger or smaller than the starting value? Volume fell, so pressure must rise — that alone eliminates 80 kPa and 80 000 kPa instantly.
• Check units are consistent before substituting. Here kPa and cm³ both cancel, so the answer comes out in kPa directly.
• Keep the unit attached to the numerical answer; an answer without a unit is incomplete even in a multiple-choice context.

Walkthrough

The diagram shows three boxes representing the states of matter: gas, liquid, and solid. We need to identify the change of state for each arrow based on its direction.

• Arrow W points from liquid to gas. The change of state from a liquid to a gas is called boiling (or evaporation).
• Arrow X points from gas to liquid. The change of state from a gas to a liquid is called condensation.
• Arrow Y points from solid to liquid. The change of state from a solid to a liquid is called melting.
• Arrow Z points from liquid to solid. The change of state from a liquid to a solid is called freezing.

Matching these to the given options:

• W = boiling
• X = condensation
• Y = melting
• Z = freezing

This exactly matches row D.

Key Takeaways

• Know the names of the changes of state between solid, liquid, and gas.
• Always check the direction of the arrow in a state change diagram, as the reverse process has a different name (e.g., melting vs. freezing).

Common Mistakes

• Reading the arrow direction the wrong way around (e.g., assuming W is gas to liquid instead of liquid to gas).
• Confusing similar-sounding terms like melting and freezing, or boiling and condensation.

Things to Be Careful About

• Boiling and evaporation both describe the change from liquid to gas, but "boiling" is the specific term used in the options provided.
• Ensure you match each letter (W, X, Y, Z) to the correct arrow and direction before selecting the row.

Walkthrough

The question asks for the mechanism of thermal energy transfer from the hotplate to the metal pan. The diagram shows the metal pan resting directly on the hotplate. Both are solids in direct physical contact. In solids, thermal energy is transferred by conduction: the particles in the hotter region (the hotplate) vibrate more vigorously and collide with neighbouring particles (in the pan), passing on kinetic energy. Therefore, the transfer is by conduction.

• Convection (Option B) is the transfer of thermal energy by the bulk movement of a fluid (liquid or gas). It does not occur between two solid surfaces in contact.
• Diffraction (Option C) and refraction (Option D) are phenomena associated with waves (such as light or sound), not with the transfer of thermal energy.

Key Takeaways

• Conduction is the transfer of thermal energy through a material or between materials in direct contact, without the material itself moving. It is the primary method of heat transfer in solids.
• Convection occurs in fluids (liquids and gases) due to density differences causing fluid movement.
• Diffraction and refraction are wave behaviours and are not mechanisms for thermal energy transfer.

Common Mistakes

• Choosing convection because the pan contains water. The question specifically asks about the transfer from the hotplate to the metal pan, not from the pan to the water. (The transfer from the pan to the water would also involve conduction at the boundary, followed by convection within the water).
• Selecting diffraction or refraction due to a lack of familiarity with thermal energy transfer terminology, confusing wave properties with thermal processes.

Things to Be Careful About

• Read the question carefully to identify exactly which two regions are involved in the energy transfer. "From the hotplate to the metal pan" means solid-to-solid contact = conduction. If it asked "from the water to the air above it", the answer would be evaporation/convection/radiation depending on context, but definitely not conduction through a solid.
• The options include wave phenomena (diffraction, refraction) to test whether the candidate knows the difference between thermal energy transfer mechanisms and wave properties. Always match the physical process (thermal transfer) to the correct category of physics.

Walkthrough

When the water at the base of the beaker is heated, the water molecules gain kinetic energy and move further apart. This causes the volume of the heated water to increase while its mass remains the same, so its density decreases. The less dense, warmer water rises upwards, and the cooler, denser water moves down to take its place. This circulation of fluid is called a convection current. Therefore, the density of the heated water decreases and it moves upwards. This matches option B.

Key Takeaways

• Heating a liquid causes it to expand and its density to decrease.
• Less dense fluid rises above denser, cooler fluid, creating convection currents.

Common Mistakes

• Thinking that heated water becomes denser (confusing mass and volume changes).
• Thinking that heated water stays in place (forgetting that convection requires movement).

Things to Be Careful About

• Convection only occurs in fluids (liquids and gases) because the particles are free to move. It does not occur in solids.
• Density is mass per unit volume. When heated, volume increases, so density decreases.

Walkthrough

The question provides a graph of pressure variation against distance for a sound wave. Sound waves are longitudinal waves, meaning the particles oscillate parallel to the direction of wave travel. This creates alternating regions of high pressure (compressions) and low pressure (rarefactions).

• Point P is at a peak (crest) above the atmospheric pressure line. This represents a region where the pressure is higher than normal, which is a compression.
• Point Q is at a trough (valley) below the atmospheric pressure line. This represents a region where the pressure is lower than normal, which is a rarefaction.
• The wavelength ($\lambda$) is defined as the distance between two consecutive points that are in phase. On this graph, the distance X is measured horizontally between two consecutive crests (compressions), so X represents one full wavelength.
• The distance Y is measured vertically from the atmospheric pressure line to a crest, representing the amplitude of the pressure variation, not the wavelength.

Matching these findings to the options: rarefaction is Q, and wavelength is X. This corresponds to option C.

Key Takeaways

• Sound waves are longitudinal waves with compressions and rarefactions.
• On a pressure-distance graph, compressions are crests above atmospheric pressure and rarefactions are troughs below it.
• Wavelength is the distance between consecutive in-phase points (e.g., crest to crest), not the amplitude.

Common Mistakes

• Confusing a pressure-distance graph with a displacement-distance graph: on a displacement graph, the points of maximum displacement are not compressions or rarefactions (they are at atmospheric pressure). On a pressure graph, a crest (high pressure) is a compression and a trough (low pressure) is a rarefaction.
• Reading Y as the wavelength: Y is the vertical distance from atmospheric pressure to a crest, which is the amplitude of the pressure variation, not the wavelength. Wavelength is a horizontal distance.

Things to Be Careful About

• Ensure you are reading the correct axis: the vertical axis is pressure, not displacement. A crest on a pressure graph is a compression, not a rarefaction.
• Wavelength is a horizontal distance (distance between consecutive compressions), not a vertical distance (amplitude).

Walkthrough

The problem asks for the speed of a wave given a graph of the cork's height against time. This graph is a displacement-time graph for the motion of a single point (the cork), not a snapshot of the wave in space.

First, read the amplitude from the graph. The height oscillates between +2.0 cm and -2.0 cm. The amplitude is the maximum displacement from the equilibrium (average) position, which is 2.0 cm.

The question gives a specific condition: 'The numerical value of the wavelength of this wave is equal to the numerical value of the amplitude of this wave.' Since the amplitude is 2.0 cm, its numerical value is 2.0. Therefore, the wavelength $\lambda$ must be 2.0 cm. Note that we are equating numerical values, so the units are handled separately in the calculation.

Next, read the time period $T$ from the graph. The wave repeats itself every 0.5 s (from t=0 to t=0.5 is one full cycle). Thus, $T = 0.5$ s.

Calculate the frequency $f$ using $f = 1/T$:

Finally, calculate the wave speed $v$ using the wave equation $v = f \lambda$:

This matches option C.

Key Takeaways

• A displacement-time graph shows how the displacement of a single point varies with time. The vertical axis gives amplitude and the horizontal axis gives time period.
• The wave speed equation $v = f \lambda$ links speed, frequency, and wavelength.
• 'Numerical value' means ignoring units when setting up an equality, but units must be included in the final calculation.

Common Mistakes

• Confusing amplitude with wavelength on the graph: The graph is height vs time, so the horizontal distance between peaks is the time period, not the wavelength. The wavelength is not shown on this graph; it is given by the text condition.
• Reading peak-to-peak as amplitude: The distance from +2.0 to -2.0 is 4.0 cm, but the amplitude is half of that, 2.0 cm. If a candidate used 4.0 as the amplitude, they would get $\lambda = 4.0$ cm and $v = 8.0$ cm/s (Option D), which is a distractor.
• Confusing period and frequency: Using $T=0.5$ directly as frequency would give $v = 0.5 \times 2.0 = 1.0$ cm/s (Option A).
• Unit errors: Ensuring the final speed is in cm/s consistent with the given wavelength in cm and frequency in Hz (1/s).

Things to Be Careful About

• Graph type: Distinguish between a displacement-distance graph (where horizontal axis is distance and peak-to-peak is wavelength) and a displacement-time graph (where horizontal axis is time and peak-to-peak is period). This is a displacement-time graph.
• Numerical value: The question carefully says 'numerical value'. Amplitude is 2.0 cm. Numerical value is 2.0. Wavelength is therefore 2.0 cm. If a student thought wavelength was 2.0 m, they would get a different answer. Always keep track of units.
• Significant figures: The graph readings (2.0, 0.5) suggest 2 significant figures. The answer 4.0 is appropriate.

Walkthrough

When a water wave moves into shallow water, the bottom of the tank affects the wave and makes it travel more slowly. So the speed of the wave decreases.

The frequency of a wave is fixed by the source that produces it. The source is not changing, so the frequency stays the same when the wave passes from deep water to shallow water.

The wave equation is

where $v$ is the speed, $f$ is the frequency and $\lambda$ is the wavelength. If the speed decreases and the frequency stays the same, then the wavelength must also decrease. So the correct option is D.

Key Takeaways

• Water waves travel more slowly in shallow water than in deep water.
• The frequency of a wave is set by the source and does not change when the wave enters a different medium.
• The wave equation $v = f \lambda$ links speed, frequency and wavelength.
• When a wave slows down, its wavelength becomes shorter.

Common Mistakes

• Choosing A, B or C by guessing that speed and wavelength change in opposite directions. Use $v = f \lambda$ instead.
• Thinking that the frequency changes when the wave enters shallow water. It does not; the source controls the frequency.
• Confusing the change in wavelength with the change in direction of the wavefronts.

Things to Be Careful About

• Both the speed and the wavelength decrease, while the frequency stays the same.
• The wave equation applies to all types of waves, including water waves, sound waves and light waves.
• In a ripple tank, a change of depth causes a change in speed, and the frequency is unchanged because it is fixed by the wave source.

Walkthrough

The question asks for the angle of incidence and the angle of reflection for a ray hitting a plane mirror. The diagram shows the angle between the incident ray and the mirror surface is $40^\circ$.

By definition, the angle of incidence is the angle between the incident ray and the normal to the surface at the point of incidence. The normal is an imaginary line perpendicular ($90^\circ$) to the mirror.

The law of reflection states that the angle of reflection is equal to the angle of incidence. Therefore:

Looking at the options, D gives $50^\circ$ for both angles. Option A incorrectly uses the $40^\circ$ angle with the mirror. Option B mixes the two values. Option C has the correct angle of incidence but incorrectly assigns a different value to the angle of reflection.

Key Takeaways

• The angle of incidence is always measured from the normal, not from the mirror surface.
• The normal is perpendicular ($90^\circ$) to the reflecting surface.
• The law of reflection states that the angle of reflection equals the angle of incidence.

Common Mistakes

• Measuring from the mirror: Students often read the $40^\circ$ angle directly as the angle of incidence. This is incorrect because angles in reflection are always measured from the normal. This leads to option A or B.
• Forgetting the law of reflection: Some students might calculate the correct angle of incidence ($50^\circ$) but then fail to realize that the angle of reflection must be the same, leading to option C.

Things to Be Careful About

• Always draw the normal: When solving reflection problems, mentally (or physically) draw the normal line perpendicular to the mirror surface at the point where the ray hits. This makes it clear that the angle given ($40^\circ$) and the angle of incidence ($50^\circ$) are complementary and add up to $90^\circ$.
• Definitions matter: The angle of incidence is strictly defined as the angle between the incident ray and the normal. There is no such thing as an "angle with the mirror" in the formal law of reflection, though it is a useful intermediate value to calculate.

Walkthrough

The critical angle $c$ is defined as the angle of incidence in a denser medium (glass) for which the angle of refraction in the less dense medium (air) is exactly 90°. When light is incident at the critical angle, the refracted ray travels along the boundary between the two media, perpendicular to the normal. At the same time, a portion of the light is partially reflected back into the glass, with the angle of reflection equal to the angle of incidence $c$. Diagram D correctly shows the refracted ray along the surface and a reflected ray back into the glass. Diagrams A, B, and C show incorrect refraction angles for the critical angle.

Key Takeaways

• The critical angle is the angle of incidence in the denser medium that produces a 90° angle of refraction.
• At the critical angle, the refracted ray travels along the boundary, and partial reflection also occurs.
• Total internal reflection only occurs for angles of incidence strictly greater than the critical angle.

Common Mistakes

• Confusing the critical angle with the angle for total internal reflection: at the critical angle, there is still a refracted ray along the boundary, whereas for angles greater than the critical angle, there is no refracted ray at all.
• Assuming the ray passes straight through without bending: light only passes straight through without refraction if it is incident along the normal (0° angle of incidence).
• Forgetting the reflected ray: even at the critical angle, partial reflection occurs, so both a refracted ray along the boundary and a reflected ray should be present in the diagram.

Things to Be Careful About

• Ensure you identify the denser medium correctly; the critical angle is measured from the normal in the denser medium (glass), not the less dense medium (air).
• Remember that the angle of refraction at the critical angle is 90° to the normal, meaning the ray travels parallel to the boundary surface, not at an angle into the air.
• Distinguish between the critical angle (refracted ray along the boundary) and angles greater than the critical angle (total internal reflection with no refracted ray).

Walkthrough

The question asks for the magnification of the image formed by a converging lens when the object is placed between the lens and its principal focus. This is the classic setup for a magnifying glass.

1. Read values from the diagram: The grid squares are 1 cm by 1 cm. The principal focus $F$ is 6 cm from the lens centre, so the focal length is $f = 6$ cm. The object $O$ is 4 cm from the lens, so the object distance is $u = 4$ cm. The object height is given as 2 cm, though we don't strictly need it if we use the distance formula for magnification.

2. Apply the lens formula: For a converging lens forming a virtual image, we use $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$, taking $u$ and $f$ as positive and allowing $v$ to be negative for a virtual image.

   $$\frac{1}{6} = \frac{1}{4} + \frac{1}{v}$$

   Rearranging for $\frac{1}{v}$:

   $$\frac{1}{v} = \frac{1}{6} - \frac{1}{4} = \frac{2}{12} - \frac{3}{12} = -\frac{1}{12}$$

   So $v = -12$ cm. The image is virtual and located 12 cm from the lens on the same side as the object.

3. Calculate magnification: Linear magnification $m$ is the ratio of image distance to object distance (magnitudes) or image height to object height.

   $$m = \frac{|v|}{u} = \frac{12}{4} = 3.0$$

   The image is upright and 3 times the size of the object (6 cm tall).

4. Match to options: The calculated magnification is 3.0, which corresponds to option C.

Key Takeaways

• When an object is placed between a converging lens and its principal focus ($u < f$), the image is virtual, upright, and magnified.
• The lens formula $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ can be used with sign conventions (virtual image distance $v$ is negative) or by rearranging to find magnification directly as $m = \frac{f}{f - u}$.
• Magnification is a dimensionless ratio; it does not require the object height unless you are calculating the image height.

Common Mistakes

• Sign errors: Forgetting that $v$ is negative for a virtual image and getting $\frac{1}{v} = \frac{1}{6} + \frac{1}{4}$, which gives an incorrect positive $v$ and wrong magnification.
• Confusing $u$ and $v$: Using the wrong distance in the magnification ratio.
• Using the wrong formula: Attempting to use the mirror formula or forgetting that magnification is $v/u$ and not $u/v$.

Things to Be Careful About

• Sign conventions: 5054 accepts the convention where $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ with $v$ negative for virtual images. Be consistent.
• Reading the diagram: Ensure you count the grid squares correctly. The focus is at 6 cm, not 5 or 7. The object is at 4 cm.
• Magnification definition: Magnification is $\frac{\text{image height}}{\text{object height}}$ or $\frac{\text{image distance}}{\text{object distance}}$. It is not the ratio of object to image.

Walkthrough

Start with the order of colours in the visible spectrum: red, orange, yellow, green, blue, indigo, violet. As you move from red to violet, the frequency $f$ increases and the wavelength $\lambda$ decreases. Therefore red light has the lowest frequency and the longest wavelength in the visible spectrum. Violet light has the highest frequency and the shortest wavelength.

Now test each option. Option A says a prism deviates red light more than violet light. This is wrong because violet light is deviated more than red light when white light passes through a prism. Option B says red light has a lower frequency than violet light. This is correct. Option C says red light has a shorter wavelength than violet light. This is wrong because red light has a longer wavelength. Option D says the speed of red light in a vacuum is less than the speed of violet light in a vacuum. This is wrong because all electromagnetic waves, including all colours of light, travel at the same speed $c$ in a vacuum.

Key Takeaways

• The visible spectrum is ordered red, orange, yellow, green, blue, indigo, violet.
• Frequency increases from red to violet.
• Wavelength decreases from red to violet.
• All electromagnetic waves travel at the same speed in a vacuum.
• A prism disperses white light because different colours are refracted by different amounts; violet is deviated most.

Common Mistakes

• Thinking red light has a shorter wavelength than violet light. In fact red has the longest wavelength in the visible spectrum.
• Thinking a prism deviates red light more than violet light. Violet is deviated more.
• Thinking different colours of light travel at different speeds in a vacuum. They all travel at $c$.

Things to Be Careful About

• Distinguish frequency from wavelength: they are inversely related for a fixed speed.
• The speed of light in a vacuum is the same for all electromagnetic waves; differences in speed in a medium cause dispersion.
• When asked about red and violet light, remember the order of the spectrum and which end has higher frequency and which end has longer wavelength.

Walkthrough

Ultrasound is sound with a frequency higher than the human ear can hear, above about 20 000 Hz. It can be used for cleaning, as stated in the question, and also for medical imaging.

Prenatal scanning uses ultrasound to produce images of a developing baby inside the mother. The ultrasound waves are sent into the body, reflected at boundaries between different tissues, and the reflected waves are used to build up an image. This is a well-known use of ultrasound, so option B is correct.

The other options are not uses of ultrasound:

• Optical fibre communication uses light travelling through thin glass fibres by total internal reflection.
• Photography uses light to form an image on a sensor or film.
• Telephone communications use electrical signals, and often light or radio waves, not sound waves.

So the correct answer is B.

Key Takeaways

• Ultrasound is sound, not an electromagnetic wave.
• Ultrasound has many uses, including cleaning, medical scanning, and sonar.
• Prenatal scanning is a medical use of ultrasound.
• Light-based technologies such as optical fibres and photography are not uses of ultrasound.

Common Mistakes

• Choosing optical fibre communication because it also involves waves. Optical fibres use light, not ultrasound.
• Thinking that any wave technology must be ultrasound. Ultrasound is specifically sound above the audible range.

Things to Be Careful About

• Remember that ultrasound needs a medium to travel through, whereas light and radio waves do not.
• In questions about uses of waves, check whether the wave is sound or electromagnetic before choosing an answer.

Walkthrough

The pulse of sound travels downwards, reflects from a surface, and travels back up. So the time given is the time for the sound to go there and come back — twice the distance we want.

For the sea bed, the round-trip time is $1.0\ \text{s}$, so the one-way time is $0.50\ \text{s}$. Using

with $v = 1500\ \text{m/s}$ gives

So the sea is $750\ \text{m}$ deep.

For the whale, the round-trip time is $0.60\ \text{s}$, so the one-way time is $0.30\ \text{s}$. The whale is therefore at a depth of

below the surface.

The whale is between the surface and the sea bed, so its distance above the sea bed is

This is option A.

Key Takeaways

• In echo and sonar questions, the given time is always the round-trip time. Halve it before calculating distance.
• Use $d = vt$ with consistent units: speed in $\text{m/s}$, time in $\text{s}$, distance in $\text{m}$.
• Read the question carefully: it asks for the distance of the whale above the sea bed, not the depth of the whale below the surface.

Common Mistakes

• Using the full round-trip times without halving them. That would give $1500\ \text{m}$ and $900\ \text{m}$, which are not the correct depths.
• Choosing $450\ \text{m}$ (option B). This is the whale's depth below the surface, not its distance above the sea bed.
• Choosing $750\ \text{m}$ (option D). This is the depth of the sea, not the distance asked for.
• Forgetting to subtract the whale's depth from the sea depth.

Things to Be Careful About

• The pulse reflects, so the time measured is for the sound to travel down and back up.
• Keep the units consistent: $1500\ \text{m/s}$ and times in seconds give distances in metres.
• Distinguish clearly between "below the surface" and "above the sea bed". The whale is closer to the surface than the sea bed, so its distance above the sea bed is smaller than the sea depth.

Walkthrough

Ultrasound is simply sound that is too high-pitched for humans to hear. The normal audible range for a healthy young person is roughly $20\ \text{Hz}$ to $20\ \text{kHz}$, so any sound with a frequency greater than $20\ \text{kHz}$ is called ultrasound.

Sound waves are longitudinal waves: the particles of the medium vibrate backwards and forwards along the same line as the direction in which the wave travels. Transverse waves, such as light waves, vibrate at right angles to the direction of travel. So the correct row must say both that ultrasound is longitudinal and that its frequency is greater than $20\ \text{kHz}$. That is row B.

Key Takeaways

• Ultrasound is sound with a frequency greater than $20\ \text{kHz}$, above the human audible range.
• Sound waves, including ultrasound, are longitudinal waves.
• The audible range is approximately $20\ \text{Hz}$ to $20\ \text{kHz}$.

Common Mistakes

• Choosing C or A because the frequency is remembered but the wave type is wrong. Sound is never transverse.
• Confusing the upper limit of the audible range with the threshold for ultrasound. Ultrasound is above $20\ \text{kHz}$, not below it.
• Mixing up the frequencies: audible sound is usually between $20\ \text{Hz}$ and $20\ \text{kHz}$, so ultrasound must be greater than $20\ \text{kHz}$.

Things to Be Careful About

• The unit $\text{kHz}$ means thousands of hertz, so $20\ \text{kHz} = 20\,000\ \text{Hz}$.
• Remember the direction of vibration for longitudinal waves: particles move parallel to the direction of energy transfer.

Walkthrough

The question gives a definition and asks for the quantity and its unit. Read the definition carefully: "work done by a source in moving unit charge around a complete circuit." This is exactly the definition of electromotive force (e.m.f.) of a source such as a cell or battery. The e.m.f. is the total energy supplied per unit charge as the charge goes all the way around the circuit from one terminal of the source back to the other.

A source with an e.m.f. of 1 V gives each coulomb of charge 1 J of energy. So the unit is the volt (V), which is the same as one joule per coulomb (J/C).

Potential difference (p.d.) is a different quantity: it is the work done per unit charge between two points in a circuit, for example across a component such as a lamp or resistor. The definition in the question says "around a complete circuit", which points to e.m.f., not p.d. Also, a newton (N) is a unit of force and would never be used for either e.m.f. or p.d. Therefore option B is correct.

Key Takeaways

• e.m.f. = the energy given to each unit of charge by a source as it moves around a complete circuit.
• p.d. = the energy transferred per unit charge between two points in a circuit.
• Both e.m.f. and p.d. are measured in volts (V), so the distinguishing feature is the phrase "around a complete circuit" for e.m.f. versus "between two points" for p.d.
• A newton (N) measures force, not electrical energy per unit charge.

Common Mistakes

• Choosing D (potential difference, V): this confuses p.d. with e.m.f. The definition "around a complete circuit" is the key phrase that identifies e.m.f.
• Choosing A or C with unit N: a newton is a unit of force (from $F = ma$), not of e.m.f. or p.d. The volt is the only correct unit here.
• Thinking e.m.f. and p.d. are the same thing: they have the same unit but different meanings. e.m.f. is the energy supplied per unit charge by a source; p.d. is the energy transferred per unit charge between two points.

Things to Be Careful About

• Always match both columns: the quantity and its unit must both be correct. Options A and C are wrong on the unit alone, even though the quantity in A is right.
• Remember the definition of e.m.f. uses the phrase "around a complete circuit" — this is the standard 5054 wording. If the definition says "between two points", it is p.d.
• The unit of e.m.f. is the volt; you do not need to quote J/C, although it is equivalent.

Walkthrough

1. Identify the circuit structure: The circuit has a $6.0\text{ V}$ d.c. supply. Starting from the positive terminal, the current flows through a $3.0\ \Omega$ resistor. Then it splits into two parallel branches, each containing a $6.0\ \Omega$ resistor. These branches recombine, and the current then flows through a third $6.0\ \Omega$ resistor (the one on the right vertical branch) before returning to the negative terminal via the switch. The voltmeter is connected across this rightmost $6.0\ \Omega$ resistor.
2. Calculate equivalent resistance of the parallel section: Two $6.0\ \Omega$ resistors in parallel give: $$R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{6.0 \times 6.0}{6.0 + 6.0} = 3.0\ \Omega$$
3. Calculate total resistance: The $3.0\ \Omega$ resistor, the $3.0\ \Omega$ parallel equivalent, and the rightmost $6.0\ \Omega$ resistor are all in series: $$R_{\text{total}} = 3.0 + 3.0 + 6.0 = 12.0\ \Omega$$
4. Calculate total current: Using the supply voltage and total resistance: $$I_{\text{total}} = \frac{V_{\text{supply}}}{R_{\text{total}}} = \frac{6.0}{12.0} = 0.5\text{ A}$$
5. Find the voltmeter reading: The voltmeter is in parallel with the rightmost $6.0\ \Omega$ resistor. The full total current ($0.5\text{ A}$) passes through this resistor. Using $V = IR$: $$V_{\text{reading}} = 0.5 \times 6.0 = 3.0\text{ V}$$

Key Takeaways

• When combining resistors, simplify parallel sections first before adding series resistances.
• In a series circuit, the current is the same through all components, so the total current flows through any single series resistor.
• A voltmeter measures the potential difference across the component it is connected in parallel with.

Common Mistakes

• Ignoring the parallel combination: Treating all resistors as being in series or calculating the parallel resistance incorrectly (e.g., $6.0 + 6.0 = 12.0\ \Omega$ instead of $3.0\ \Omega$).
• Misreading the voltmeter connection: Thinking the voltmeter is in series with a resistor or measuring the total supply voltage. The voltmeter is clearly in parallel with only the rightmost $6.0\ \Omega$ resistor.
• Calculating current through the wrong branch: Dividing the current by 2 for the parallel branches is unnecessary here because the voltmeter is across a resistor in the main series path, not one of the parallel branches.

Things to Be Careful About

• Always simplify the circuit diagram step-by-step: parallel parts first, then series parts.
• Check exactly what the voltmeter is connected across. In this case, it is across a single $6.0\ \Omega$ resistor that carries the full circuit current, not across the parallel combination.
• Units: ensure voltage is in volts, resistance in ohms, and current in amperes to get the correct result.

Walkthrough

1. Identify the circuit type: The diagram shows two resistors $R_1$ and $R_2$ connected in series.
2. Recall the series circuit rule: In a series circuit, there is only one path for the current, so the current $I$ is the same through every component.
3. Apply Ohm's law: For each resistor, the potential difference is given by $V = IR$. Therefore, $V_1 = I R_1$ and $V_2 = I R_2$.
4. Form the ratio: To find $V_1 / V_2$, divide the first equation by the second:

The current $I$ cancels out, leaving:

This corresponds to option A.

Key Takeaways

• In a series circuit, the current is constant throughout.
• The potential difference across a resistor is directly proportional to its resistance when the current is constant.
• A potential divider splits the total voltage in the exact ratio of the resistances: $V_1 / V_2 = R_1 / R_2$.