Physics 5054/12 — October/November 2024

Walkthrough

The question asks for the most suitable device to obtain an accurate value for the height of the clothes peg, where the height is indicated as the small vertical gap between the front jaws of the peg.

We evaluate each option:

• A tape: Tape measures are flexible and generally have 1 mm graduations. They are suitable for longer, less precise measurements (like the length of a room) but not for accurate small gaps.
• A metre rule: A standard metre rule has 1 mm graduations. While it can measure small lengths, it is not as accurate as a micrometer for very small distances.
• A micrometer: Also known as a micrometer screw gauge, this instrument is specifically designed to measure small lengths (such as the thickness of a wire or a small gap) accurately to 0.01 mm. It is the most precise tool listed for a small linear measurement.
• A measuring cylinder: This is an instrument used to measure the volume of liquids, not a linear dimension like length or height.

Since the question asks for an accurate value for a small height, the micrometer is the correct choice.

Key Takeaways

• Different measuring instruments have different ranges and precisions. Choose the instrument that matches both the size of the quantity and the required accuracy.
• Micrometers (screw gauges) are used for small, precise linear measurements (e.g., thickness of wire, small gaps).
• Metre rules and tapes are used for larger, less precise measurements.
• Measuring cylinders are for volume, not length.

Common Mistakes

• Choosing a metre rule simply because it can measure length, without considering that the question asks for an accurate value for a small gap, which requires a more precise instrument like a micrometer.
• Choosing a tape because it is commonly used for measuring clothes or pegs, but ignoring its lack of accuracy.
• Choosing a measuring cylinder by misreading "height" as a volume or confusing it with a measuring device.

Things to Be Careful About

• Always read the diagram carefully to see exactly which dimension is being asked for. Here, the "height" is the small vertical gap between the jaws, not the overall length of the peg (which would be around 10 cm and could be measured with a metre rule).
• Pay attention to the word accurate. If a question asks for an accurate measurement of a small length, a micrometer or vernier caliper is required, not a metre rule.

Walkthrough

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

Test each option:

• A — acceleration, displacement and velocity are all directed quantities. Displacement is a distance in a stated direction, velocity is a speed in a stated direction, and acceleration is a change in velocity, so it too has a direction. All three are vectors.
• B — chemical energy, mass and power have no direction. Energy and power are scalars; mass is a scalar. All three are scalars.
• C — force is a vector, but extension (a change in length) and gravitational potential energy (an energy) are scalars. Mixed, so not the answer.
• D — weight is a vector (a force acting downwards), but kinetic energy and work are scalars. Mixed, so not the answer.

Only option A contains quantities that are all vectors.

Key Takeaways

• A vector has both magnitude and direction; a scalar has magnitude only.
• Common vectors: displacement, velocity, acceleration, force, weight, momentum.
• Common scalars: distance, speed, mass, time, temperature, energy, work, power.
• This question tests the ability to sort everyday quantities into the two classes quickly and correctly.

Common Mistakes

• Confusing weight (a force, so a vector) with mass (a scalar).
• Confusing displacement (vector) with distance (scalar), and velocity (vector) with speed (scalar).
• Thinking that because work = force × distance, work must be a vector — it is not; work and all forms of energy are scalars.
• Thinking that power is a vector because it involves force — power is a scalar.

Things to Be Careful About

• Force is a vector, but the quantities derived from it are not automatically vectors: work and pressure are scalars.
• Extension is a change in length, so it is a scalar even though it is caused by a force.
• Acceleration is a vector even in straight-line motion; its direction is the direction of the change in velocity.

Walkthrough

1. Identify the goal: The question asks for the average speed of the cyclist for the entire ride. Average speed is a global property of the trip, not the speed at any particular moment.
2. Recall the formula: The formula for average speed is: $$\text{average speed} = \frac{\text{total distance}}{\text{total time}}$$
3. Read the total time: The question states the ride lasts for $14\text{ s}$. We can also verify this on the graph; the horizontal axis (time) goes from 0 to 15, and the graph line ends at $t = 14\text{ s}$ (one major grid unit before 15, where each major unit is 1 s).
4. Read the total distance: Look at the vertical axis (distance) at the end of the graph line ($t = 14\text{ s}$). The vertical axis has major markings at 0, 10, 20, 30. There are 5 large squares between 0 and 10, so each large square represents $2\text{ m}$. The graph ends 4 large squares below 30, which is $30 - 8 = 22$? No, let's look closer. 20 is a line. 4 squares above 20 is $20 + (4 \times 2) = 28\text{ m}$. The graph ends at $28\text{ m}$.
5. Calculate: Substitute the values into the formula: $$\text{average speed} = \frac{28}{14} = 2.0\text{ m/s}$$
6. Match with options: The calculated value is $2.0\text{ m/s}$, which corresponds to option A.

Key Takeaways

• Average speed depends only on the total distance and total time for the whole journey. The details of the motion (stopping, speeding up, slowing down) do not matter for the average speed calculation.
• Distance-time graphs: The vertical axis gives the total distance from the start at any given time. The gradient (slope) at any point gives the instantaneous speed at that time.

Common Mistakes

• Calculating the speed of a specific section: Candidates might calculate the speed during the final moving phase (from $t=6\text{ s}$ to $t=14\text{ s}$). Distance = $28 - 8 = 20\text{ m}$, Time = $14 - 6 = 8\text{ s}$, Speed = $20/8 = 2.5\text{ m/s}$. This matches option D, but it is the speed during the last part, not the average speed for the whole ride.
• Reading the graph incorrectly: Misreading the final distance value (e.g., reading 32 m or 30 m) would lead to incorrect answers like B or C.
• Confusing distance and displacement: In 1D motion without turning back, distance equals the magnitude of displacement. Here, the cyclist moves forward, stops, then moves forward again. Total distance is simply the final y-value.

Things to Be Careful About

• Graph reading precision: Ensure you read the grid correctly. The x-axis has 5 large squares per 5 seconds (1 s per large square). The y-axis has 5 large squares per 10 meters (2 m per large square). The final point is at $(14, 28)$.
• Definition of average speed: Do not average the speeds of different sections (e.g., speed in first part + speed in second part divided by 2). That is a common mathematical error. Always use $\text{total distance} / \text{total time}$.
• Units: Ensure the final answer is in $\text{m/s}$. The graph axes are in meters and seconds, so the result is naturally in $\text{m/s}$.

Walkthrough

The question asks for sections where the magnitude of the velocity is decreasing. The magnitude of velocity is speed. On a velocity-time graph, speed is the absolute value of the velocity, meaning we look for sections where the graph is moving towards the time axis (velocity = 0).

• In section P, the velocity goes from a positive value down to 0. The speed is decreasing.
• In section Q, the velocity goes from 0 down to a negative peak. The speed (magnitude) is increasing.
• In section R, the velocity goes from the negative peak back up to 0. The speed (magnitude) is decreasing.
  Thus, P and R are the correct sections.

Key Takeaways

• Velocity is a vector; its magnitude is speed.
• On a velocity-time graph, decreasing speed corresponds to the graph moving towards the time axis (v = 0), regardless of whether the velocity is positive or negative.

Common Mistakes

• Confusing 'velocity decreasing' with 'speed decreasing'. Velocity is decreasing throughout P, Q, and R (the gradient is negative everywhere), but speed only decreases in P and R.
• Forgetting that a negative velocity with an increasing magnitude (becoming more negative) means speed is increasing.

Things to Be Careful About

• Always read 'magnitude of velocity' as 'speed'.
• Check the sign of the velocity: if it is negative and moving away from zero, the magnitude is increasing.

Walkthrough

The question asks which property of a metal disc remains unchanged when it is squeezed in a vice. We evaluate each option using the definition of mass and the kinetic particle model of solids.

• Option A (mass): Mass is a measure of the amount of matter in an object. Squeezing the disc deforms it but does not add or remove any metal atoms. Therefore, the mass is conserved and unchanged.
• Option B (radius): As the circular disc is squeezed, it flattens into an oval or elliptical shape. The radius in the direction of the squeezing force decreases, while the radius in the perpendicular direction increases. Thus, the radius changes.
• Option C (distance between its atoms): In a solid, atoms are arranged in a regular lattice. When squeezed, the lattice is distorted. The atoms move closer together in the direction of the applied force and spread further apart in the perpendicular direction. The distances between atoms therefore change.
• Option D (forces between its atoms): Atoms in a solid are held at an equilibrium separation by attractive and repulsive forces. When the disc is squeezed, atoms are pushed closer together than their equilibrium separation. This increases the repulsive forces between them, which is why the metal resists being squeezed. Thus, the forces between atoms change.

Key Takeaways

• Mass is an intrinsic property representing the amount of matter; it does not change with shape, volume, or position.
• Deforming a solid changes its macroscopic dimensions (like radius) and its microscopic atomic arrangement (distances and forces between atoms), but not the total amount of matter.

Common Mistakes

• Thinking that mass changes when the shape of an object changes. Mass is conserved during deformation.
• Assuming that the distance between atoms in a solid is fixed. While solids are rigid, their atomic lattice can be distorted under force, changing interatomic distances.
• Confusing mass with weight. While weight is also unchanged here (assuming $g$ is constant), it is not an option, and mass is the fundamental property being tested.

Things to Be Careful About

• Ensure you distinguish between macroscopic properties (radius, shape) and microscopic properties (atomic distances, intermolecular forces). Both can change during deformation.
• Remember that 'mass' refers to the quantity of matter, not the volume or density. Volume and density will change as the shape changes (if the volume is conserved, density is constant; if volume changes due to compression, density changes, but mass remains constant).

Walkthrough

Mass is a property of the object itself: it is the amount of matter in the satellite, so it stays the same wherever the satellite is. Weight is the gravitational force acting on the object, and it is calculated using $W = mg$, where $m$ is the mass and $g$ is the gravitational field strength.

On the Earth's surface, $g$ is about 10 N/kg. In space, the gravitational field strength is smaller, so the same mass gives a smaller weight. Therefore the correct answer is D: the mass is unaffected and the weight decreases.

Option A is wrong because the gravitational field strength is smaller in space, so the weight cannot be unaffected. Option B is wrong because mass does not depend on gravity. Option C is wrong because mass does not increase and weight is not unaffected.

Key Takeaways

• Mass is the amount of matter in an object and is measured in kilograms.
• Weight is a force, measured in newtons, and equals the mass multiplied by the gravitational field strength.
• Mass stays the same everywhere; weight changes when the gravitational field strength changes.
• A satellite in space still has weight, but it is smaller than on the Earth's surface.

Common Mistakes

• Thinking that mass decreases in space. Mass is not affected by gravitational field strength.
• Confusing mass with weight and choosing an option that treats them as the same thing.
• Forgetting that weight is a force, so it depends on the value of $g$.

Things to Be Careful About

• The question says the gravitational field strength is smaller, not zero, so the weight decreases but is not necessarily zero.
• Mass is measured in kg, while weight is measured in N.
• When using $W = mg$, use the gravitational field strength in N/kg and mass in kg.

Walkthrough

This is a unit conversion problem. The density is given as $2.4\text{ g/cm}^3$, and we must express it in $\text{kg/m}^3$. The key is to convert the mass unit (g to kg) and the volume unit (cm³ to m³) separately.

1. Convert mass: There are 1000 grams in a kilogram, so $1\text{ g} = 1 \times 10^{-3}\text{ kg}$.
2. Convert volume: There are 100 centimetres in a metre. Since the unit is cubic, we must cube this conversion factor: $1\text{ m}^3 = (100\text{ cm})^3 = 1 \times 10^{6}\text{ cm}^3$. This means $1\text{ cm}^3 = 1 \times 10^{-6}\text{ m}^3$.
3. Combine the conversions: Substitute these into the density, treating the units as algebraic quantities. The $\text{cm}^3$ in the denominator of the density is replaced by $1 \times 10^{-6}\text{ m}^3$. This gives $2.4 \times 10^{-3} / 10^{-6} = 2.4 \times 10^{3}\text{ kg/m}^3$.
4. Final answer: $2.4 \times 10^{3} = 2400$, so the density is $2400\text{ kg/m}^3$. This matches option D.

Key Takeaways

• Unit conversion is a core skill: This question tests the ability to convert between units of density, which is a fundamental physical property.
• Cubic units: The most common mistake is forgetting to cube the linear conversion factor for volume. A factor of 100 for length becomes a factor of $100^3 = 10^6$ for volume.
• Powers of ten: Being comfortable with scientific notation and powers of ten is essential for solving these types of problems quickly and accurately.

Common Mistakes

• Forgetting to cube the conversion factor: A very common error is to convert $\text{cm}^3$ to $\text{m}^3$ using a factor of 100 instead of $100^3 = 1 \times 10^6$. This would give $2.4 \times 10^{-1} = 0.24\text{ kg/m}^3$, which is option B.
• Incorrect direction of conversion: Dividing instead of multiplying by $10^6$ would give $2.4 \times 10^{-9}\text{ kg/m}^3$, which is not an option but is a common slip.
• Not using standard form: Attempting to write out $1 \times 10^6$ as 1,000,000 can lead to errors in counting zeros, especially when dividing.

Things to Be Careful About

• Always cube the linear conversion factor when dealing with volume or area units.
• Check the order of magnitude. The density of most solids and liquids is on the order of $10^3\text{ kg/m}^3$, so an answer like $0.24\text{ kg/m}^3$ should immediately look wrong for a solid material.

Walkthrough

The object is falling vertically downwards, as indicated by the 'direction of fall' arrow in the diagrams. There are two forces acting on the object:

1. Weight (labelled 'force from Earth on object'): This is the gravitational pull of the Earth. Weight always acts vertically downwards, towards the centre of the Earth. Therefore, the arrow for this force must point down.
2. Air resistance (labelled 'force of air on object'): This is a drag force that opposes the motion of the object through the air. Since the object is moving downwards, the air resistance must act upwards. Therefore, the arrow for this force must point up.

Diagram A is the only diagram that correctly shows the force from Earth pointing down and the force of air pointing up.

Key Takeaways

• Weight is the force exerted by a gravitational field (Earth) on a mass, and it always acts downwards.
• Air resistance (or drag) is a frictional force that always acts in the direction opposite to the motion of the object through a fluid (air or liquid).
• In a free-body diagram, only forces are drawn as arrows. The direction of motion (velocity) is not a force and is not drawn as a force arrow.

Common Mistakes

• Choosing a diagram where air resistance points downwards, confusing the direction of motion with the direction of the drag force.
• Thinking that because the object is falling, all forces must point downwards.
• Including the 'direction of fall' as a force arrow in the diagram.

Things to Be Careful About

• Always distinguish between the direction of motion (velocity) and the direction of forces. Forces cause changes in motion, they do not necessarily point in the direction of motion.
• Remember that weight is a force towards the Earth, so it is always downwards regardless of whether the object is moving up, down, or sideways.

Walkthrough

The question asks for the expression for the moment of force $F$ about the pivot. The fundamental definition of the moment of a force is the product of the applied force and the perpendicular distance from the pivot to the line of action of that force. Looking at Fig. 1, the force $F$ is applied at an angle to the lever. The distance $l$ is measured along the lever from the pivot to the point of application, but this is not the perpendicular distance. The perpendicular distance from the pivot to the line of action of $F$ is explicitly labelled as $d$. Therefore, the moment is simply $F \times d$. Option D ($F \times l$) is a common distractor for candidates who forget that the distance must be measured perpendicularly from the pivot to the line of action.

Key Takeaways

• The moment of a force is $F \times d$, where $d$ is the perpendicular distance from the pivot to the line of action of the force.
• The distance along the lever ($l$) is not used directly in the moment calculation unless the force is applied perpendicularly to the lever.

Common Mistakes

• Using the distance along the lever ($l$) instead of the perpendicular distance ($d$). This leads to choosing option D.
• Dividing the force by the distance instead of multiplying, leading to option A or B.

Things to Be Careful About

• Always check whether the given distance is perpendicular to the line of action of the force. If the force is applied at an angle, you must use the perpendicular distance from the pivot to the line of action, not the distance along the lever.

Walkthrough

This question asks for the work done to increase the rocket's speed. Work done is the product of the resultant force and the distance moved in the direction of that force. The rocket is moving upwards, so we first find the net upward force.

The forces are:

• Upward engine force: $3.5\text{ MN}$
• Downward weight: $1.9\text{ MN}$
• Downward air resistance: $0.10\text{ MN}$

The resultant force is:

Then, work done is:

So the correct answer is A.

• B (170 MJ) is the work done if you forgot to include air resistance (i.e., using 1.7 MN as the resultant force).
• C (350 MJ) is the work done by the engine force alone, ignoring weight and air resistance.
• D (550 MJ) is the work done if you added the forces instead of subtracting (3.5 + 1.9 + 0.1 = 5.5 MN).

Key Takeaways

• Work done to increase speed is calculated using the resultant force, not just any individual force.
• Always identify the direction of each force and combine them as vectors (here, all forces are along a line).
• Convert units (MN to N) before calculating.

Common Mistakes

• Forgetting to subtract air resistance – leads to option B.
• Using the engine force alone – leads to option C. The weight and air resistance oppose the motion.
• Adding all forces – leads to option D. This is incorrect because weight and air resistance act opposite to the engine force.

Things to Be Careful About

• The unit MN (meganewton) must be converted to newtons: $1\text{ MN} = 10^6\text{ N}$.
• Work is a scalar, but you must use the resultant force in the direction of motion.
• The distance is already in metres, so no conversion is needed there.

Walkthrough

Efficiency tells us how good a device or system is at transferring energy usefully. It is always a ratio of what you get out that is useful, divided by what you put in. So, efficiency = (useful energy output) / (total energy input). Option A matches this definition exactly.

Option B is the inverse, which would give a value greater than 1 (more than 100%), which is impossible for efficiency. Option C incorrectly uses power instead of energy, and Option D multiplies the two quantities, which would give incorrect units and not a ratio.

Key Takeaways

• Efficiency is a measure of how well energy is converted into a useful form.
• It is always calculated as (useful output) divided by (total input).
• Since it is a ratio, it has no units and is often expressed as a percentage.

Common Mistakes

• Choosing option B, which inverts the ratio. Remember, efficiency cannot be greater than 1 (or 100%).
• Confusing energy with power. Efficiency compares energy (or power) output to energy (or power) input, but the question specifically asks for energy transfer.

Things to Be Careful About

• Remember that efficiency is a ratio, so it has no units.
• Ensure you are comparing the same quantity (energy in this case) in the numerator and denominator.

Walkthrough

The question asks for the greatest power transferred when lifting objects. Power is defined as the rate at which work is done, given by the equation $P = W/t$. When lifting an object vertically against gravity, the work done $W$ is equal to the force applied (weight, $mg$) multiplied by the distance lifted ($d$ or $h$). So, $W = mgh$.

Substituting this into the power equation gives $P = mgh/t$. Since the gravitational field strength $g$ is constant for all four options (we can use $g = 10 \text{ N/kg}$ for calculation), the power is directly proportional to $\frac{m \times d}{t}$. We can calculate the power for each option to find the maximum.

• For A: Mass $2.0 \text{ kg}$, distance $2.0 \text{ m}$, time $3.0 \text{ s}$. Work done $= 2.0 \times 10 \times 2.0 = 40 \text{ J}$. Power $= 40 / 3.0 = 13.3 \text{ W}$.
• For B: Mass $1.5 \text{ kg}$, distance $2.5 \text{ m}$, time $2.0 \text{ s}$. Work done $= 1.5 \times 10 \times 2.5 = 37.5 \text{ J}$. Power $= 37.5 / 2.0 = 18.75 \text{ W}$.
• For C: Mass $2.0 \text{ kg}$, distance $2.0 \text{ m}$, time $1.5 \text{ s}$. Work done $= 2.0 \times 10 \times 2.0 = 40 \text{ J}$. Power $= 40 / 1.5 = 26.7 \text{ W}$.
• For D: Mass $1.5 \text{ kg}$, distance $2.5 \text{ m}$, time $1.5 \text{ s}$. Work done $= 1.5 \times 10 \times 2.5 = 37.5 \text{ J}$. Power $= 37.5 / 1.5 = 25 \text{ W}$.

The highest value is $26.7 \text{ W}$, which corresponds to option C.

Key Takeaways

• Power is the rate of doing work: $P = W/t$.
• Work done against gravity is calculated as $W = mgh$ (mass $\times$ g $\times$ height/distance).
• To compare power in different scenarios where $g$ is constant, you can compare the ratio $\frac{m \times d}{t}$.

Common Mistakes

• Confusing power with work or energy: A student might calculate the work done ($mgh$) and stop there. Option A and C have the same work done ($40 \text{ J}$), but C is faster, so it has more power. Option B and D have less work done ($37.5 \text{ J}$).
• Forgetting to divide by time: Power depends on how quickly the work is done. Option D has a shorter time than B, so it has higher power despite the same work. Option C has the shortest time combined with the highest work (tied with A), making it the winner.
• Using the wrong formula: Using $P = Fv$ might be valid if velocity was constant and calculated, but $P = mgh/t$ is more direct here.

Things to Be Careful About

• Units: Ensure mass is in kg, distance in m, and time in s to get power in Watts (J/s). The values given are already in standard units.
• Value of $g$: The mark scheme accepts calculations using $g = 9.8 \text{ N/kg}$ or $g = 10 \text{ N/kg}$. Since we are comparing ratios, the specific value of $g$ does not change the final answer, but it must be consistent if calculating exact values.
• Significant figures: The calculations result in repeating decimals (e.g., $13.333...$). Rounding to 2 or 3 significant figures is sufficient for comparison ($13.3$, $18.8$, $26.7$, $25.0$).

Walkthrough

The question states that the temperature of the trapped air remains constant while the piston is pushed down. This is a classic application of Boyle's law, which is expressed in the 5054 syllabus as $p_1 V_1 = p_2 V_2$ for a fixed mass of gas at constant temperature.
The container has a uniform cross-section, so its volume is proportional to the height of the air column: $V = A \times h$. The area $A$ cancels out when we substitute into the gas law, leaving $p_1 h_1 = p_2 h_2$.
Rearranging for the new pressure $p_2$ gives $p_2 = p_1 h_1 / h_2$.
Substituting the given values: $p_1 = 1.0 \times 10^5 \text{ N/m}^2$, $h_1 = 0.18 \text{ m}$, and $h_2 = 0.15 \text{ m}$.
$p_2 = (1.0 \times 10^5 \times 0.18) / 0.15 = 1.2 \times 10^5 \text{ N/m}^2$.
This matches option C.

Key Takeaways

• Boyle's law ($p_1 V_1 = p_2 V_2$) applies when temperature and mass are constant.
• For a container with uniform cross-section, volume is proportional to height, so $p_1 h_1 = p_2 h_2$.

Common Mistakes

• Inverting the ratio and calculating $p_2 = p_1 h_2 / h_1$, which gives $0.83 \times 10^5 \text{ N/m}^2$ (option B). Pressure must increase when volume decreases.
• Forgetting to include the $10^5$ factor and getting $1.2$ instead of $1.2 \times 10^5$.

Things to Be Careful About

• Ensure the units of height are consistent (both in metres here, so no conversion needed).
• Remember that decreasing volume at constant temperature must result in an increase in pressure; if your answer is less than the initial pressure, you have inverted the ratio.

Walkthrough

The question asks about the relationship between temperature and pressure for a gas held at constant volume. According to the kinetic particle model, gas pressure arises from the continuous collisions of gas particles with the walls of the container. When the temperature of the gas increases, the thermal energy of the particles increases, which means their average kinetic energy increases. As a result, the particles move at higher speeds. Faster-moving particles collide with the container walls more often and with greater momentum change per collision. Both of these effects (more frequent collisions and harder collisions) cause the pressure to increase. Conversely, if the temperature decreases, the particles slow down and the pressure falls. Therefore, the only row that correctly describes a valid cause-and-effect relationship is C: temperature increases, pressure increases.

Key Takeaways

• Gas pressure is caused by particle collisions with the container walls.
• Temperature is a measure of the average kinetic energy of the particles.
• At constant volume, increasing temperature increases particle speed, leading to more frequent and more forceful collisions, which increases pressure.

Common Mistakes

• Confusing the effect of volume on pressure with the effect of temperature. At constant volume, only temperature changes affect pressure in this scenario.
• Thinking that pressure remains unchanged if the container is rigid, without considering that temperature changes still alter particle speeds and collision forces.

Things to Be Careful About

• Ensure you read the question carefully to note that the volume is constant. If the volume were allowed to change (e.g., a free-moving piston), the outcome would be different.
• Remember that temperature must be in kelvin for quantitative calculations using gas laws, but the qualitative relationship (higher temperature = higher pressure at constant volume) holds regardless.

Walkthrough

The Kelvin scale is the thermodynamic temperature scale, and it is related to the Celsius scale by a simple fixed shift. A temperature difference of 1 kelvin is exactly the same size as a temperature difference of 1 degree Celsius, so the two scales only differ in where their zero sits.

Absolute zero on the Kelvin scale is $0\text{ K}$, which corresponds to $-273\,^\circ\text{C}$ on the Celsius scale. This means that to convert any Celsius temperature into kelvin you add 273:

For gold, the melting temperature is $1064\,^\circ\text{C}$, so:

That matches option D.

Each of the other options comes from a misunderstanding of the conversion. Option A ($791\text{ K}$) is $1064 - 273$, subtracting instead of adding. Option B ($964\text{ K}$) is $1064 - 100$, a made-up shift of 100. Option C ($1164\text{ K}$) is $1064 + 100$, again using 100 instead of 273. Only option D uses the correct shift of 273.

Key Takeaways

• The Kelvin and Celsius scales have the same size of degree; they differ only in their zero points.
• Absolute zero is $0\text{ K} = -273\,^\circ\text{C}$, so $T(\text{K}) = \theta(^\circ\text{C}) + 273$.
• A temperature given in kelvin is always numerically larger than the same temperature in degrees Celsius by 273.

Common Mistakes

• Subtracting 273 instead of adding it, which gives option A ($791\text{ K}$).
• Using 100 as the conversion shift instead of 273, which gives options B and C.
• Forgetting that the conversion is an addition of 273, not a multiplication or a shift of 100.

Things to Be Careful About

• Always add 273 when going from Celsius to kelvin, and subtract 273 when going from kelvin to Celsius.
• The unit for the kelvin scale is written as K with no degree symbol, so the answer is $1337\text{ K}$, not $1337\,^\circ\text{K}$.
• If a question gives a temperature in kelvin and asks for Celsius, the answer will be numerically smaller by 273.

Walkthrough

A substance is in one of three states depending on its temperature relative to its melting and boiling points:

• solid below the melting temperature,
• liquid between the melting and boiling temperatures,
• gas above the boiling temperature.

We must find the single temperature at which argon, nitrogen and oxygen are in three different states.

• At $-220\,^\circ\text{C}$, all three elements are below their melting points, so all are solid.
• At $-205\,^\circ\text{C}$, argon is solid (below $-189$), while nitrogen and oxygen are both liquid (each between its melting and boiling points). Two liquids, one solid — not three different states.
• At $-190\,^\circ\text{C}$, argon is solid (below its melting point of $-189$), nitrogen is gas (above its boiling point of $-196$), and oxygen is liquid (between $-218$ and $-183$). This gives three different states, so C is correct.
• At $-175\,^\circ\text{C}$, all three elements are above their boiling points, so all are gas.

Key Takeaways

The state of a substance depends only on where its temperature sits relative to its melting and boiling points. Comparing several temperatures against the given data, one at a time, is the reliable method.

Common Mistakes

• Confusing the direction of the inequalities: a temperature more negative than the melting point means solid, not liquid.
• Reading $-190$ as above argon's melting point of $-189$; $-190$ is actually below $-189$, so argon is solid there.
• Forgetting that a substance above its boiling point is a gas, not a liquid.

Things to Be Careful About

• All temperatures here are negative, so "higher" means less negative (e.g. $-175$ is higher than $-189$).
• Check each element separately at each option temperature before deciding.
• The correct option is the only one where the three states are all different.

Walkthrough

Evaporation is the escape of molecules from the surface of a liquid at temperatures below the boiling point. The rate of evaporation depends on four main factors:

1. Temperature: Higher temperature means molecules have more kinetic energy, so more can escape. Heating the liquid (Option B) increases the rate.
2. Surface area: A larger surface area exposes more molecules to the air. Transferring to a beaker with a larger diameter (Option D) increases the rate.
3. Air movement: Blowing air across the surface (Option A) removes the vapour molecules that have just escaped, preventing them from condensing back into the liquid, which increases the rate.
4. Vapour concentration: If the air above the liquid is saturated with vapour, evaporation slows down. Putting a lid on the beaker (Option C) traps the vapour above the liquid, increasing the vapour concentration and reducing the net rate of evaporation.

Therefore, putting a lid on the beaker is the only action that reduces the rate of loss of liquid by evaporation.

Key Takeaways

The rate of evaporation is increased by: higher temperature, larger surface area, air movement across the surface, and lower vapour concentration above the liquid. Conversely, a lid or cover reduces the rate by trapping vapour.

Common Mistakes

• Choosing Option A, thinking that blowing air cools the liquid and therefore reduces evaporation. While blowing air does increase cooling by evaporation, it actually increases the rate of evaporation itself.
• Choosing Option D, confusing surface area effects. A larger diameter means a larger surface area, which increases evaporation, not reduces it.

Things to Be Careful About

Read the question carefully: it asks what reduces the rate of loss, not what increases it. Options A, B, and D all increase the rate of evaporation. Only Option C reduces it.

Walkthrough

The question asks which surface absorbs infrared radiation best. Infrared radiation is thermal radiation, and its absorption depends on both the colour and the texture of the surface.

• Colour: Black surfaces absorb more infrared radiation than white surfaces. White surfaces reflect most of the radiation, so they stay cooler when exposed to infrared.
• Texture: Dull surfaces absorb more infrared radiation than shiny surfaces. A shiny surface reflects a large amount of the radiation away, so it is not a good absorber.

Therefore, the best absorber is black and dull, which is option B.

Key Takeaways

• The best absorber and emitter of infrared radiation is a black, dull surface.
• The best reflector of infrared radiation is a white, shiny surface.
• The same surface properties that make a good absorber also make a good emitter of infrared radiation.

Common Mistakes

• Choosing black and shiny: a shiny surface reflects some infrared radiation, so it is not the best absorber.
• Choosing white and dull: a white surface reflects most infrared radiation, regardless of texture.
• Confusing the best absorber with the best reflector: the question asks about absorption, not reflection.

Things to Be Careful About

• “Texture” here means dull/matte versus shiny, not whether the surface is rough or smooth in a general sense.
• A dull black surface is also the best emitter of infrared radiation, which is often tested alongside absorption.
• Read the options carefully: both colour and texture must be correct to score the mark.

Walkthrough

Ultrasound is defined as sound with a frequency higher than 20 kHz (20 000 Hz), which is above the human audible range. The question gives velocity and wavelength for four waves, and we need to identify the one that is ultrasound.

The wave equation links these quantities: $v = f \lambda$, so the frequency is $f = v/\lambda$. Before dividing, we must convert the wavelength from cm to m (6.6 cm = 0.066 m, 89 cm = 0.89 m, 6.0 cm = 0.060 m, 25 cm = 0.25 m).

Row A: 330 m/s is the speed of sound in air; $330/0.066 = 5000\ \text{Hz} = 5\ \text{kHz}$, which is audible, not ultrasound.

Row B: $890/0.89 = 1000\ \text{Hz}$, audible.

Row C: 1500 m/s is the speed of sound in water (and body tissue); $1500/0.060 = 25\,000\ \text{Hz} = 25\ \text{kHz}$, which is above 20 kHz, so this is ultrasound. ✓

Row D: $4000/0.25 = 16\,000\ \text{Hz} = 16\ \text{kHz}$, still audible (just below the 20 kHz threshold).

So C is the correct answer.

Key Takeaways

• Ultrasound is sound with a frequency above 20 kHz.
• The wave equation $v = f \lambda$ lets you find the frequency from the velocity and wavelength.
• The speed of sound differs in different media: about 330 m/s in air and about 1500 m/s in water (and body tissue), which is why medical ultrasound uses the 1500 m/s value.

Common Mistakes

• Forgetting to convert the wavelength from cm to m before dividing — this gives a frequency 100 times too large.
• Thinking ultrasound must travel in air at 330 m/s; in fact medical ultrasound travels in water/gel.
• Confusing wavelength with frequency, or picking a row just because the velocity looks familiar.

Things to Be Careful About

• Convert 6.0 cm to 0.060 m before using $f = v/\lambda$.
• Remember the threshold: 20 kHz = 20 000 Hz.
• A wave with frequency below 20 kHz is audible, not ultrasound, even if it has a high velocity.

Walkthrough

The problem asks for the angle of reflection for a ray hitting a plane mirror. The diagram provides the angle between the incident ray and the mirror surface ($30^\circ$). In physics, angles in reflection are always measured from the normal, which is an imaginary line drawn perpendicular ($90^\circ$) to the mirror surface at the point of incidence.

1. Find the angle of incidence ($i$): Since the normal is at $90^\circ$ to the mirror, the angle of incidence is the complement of the angle given. $i = 90^\circ - 30^\circ = 60^\circ$.
2. Apply the law of reflection: The law states that the angle of reflection ($r$) is equal to the angle of incidence ($i$). Therefore, $r = 60^\circ$.
3. Select the option: Option B matches this value.

Key Takeaways

• The angle of incidence and angle of reflection are always measured from the normal, not from the mirror surface.
• The law of reflection is $i = r$.
• If the angle with the mirror is given, subtract it from $90^\circ$ to find the angle of incidence.

Common Mistakes

• Choosing A ($30^\circ$): This is the angle between the ray and the mirror, not the angle of reflection. Students often forget to measure from the normal.
• Choosing D ($120^\circ$): This is the total angle between the incident and reflected rays ($60^\circ + 60^\circ$), not the angle of reflection itself.

Things to Be Careful About

• Always check which angle is given in the diagram. Is it with the normal or with the mirror?
• Remember that the normal is always perpendicular to the reflecting surface.

Walkthrough

The image formed by a plane mirror has four key properties: it is virtual, upright (same orientation as the object), the same size as the object, and located at the same distance behind the mirror as the object is in front of it.

• Diagram A shows the image upright, the same size as the object, and at the same distance behind the mirror. This matches all properties.
• Diagram B shows the image inverted (pointing downwards). Plane mirrors do not invert images top-to-bottom.
• Diagram C shows the image touching the mirror and inverted. Both the distance and orientation are wrong.
• Diagram D shows the image touching the mirror. The distance is wrong; the image must be behind the mirror at the same distance as the object is in front.
  Therefore, diagram A is correct.

Key Takeaways

• Plane mirror images are always upright, not inverted top-to-bottom.
• The image is the same size as the object.
• The image is located exactly as far behind the mirror as the object is in front of it.

Common Mistakes

• Assuming plane mirrors invert images top-to-bottom (like a real image from a converging lens or concave mirror). Plane mirrors only laterally invert (swap left and right), which is not visible for a symmetric upward-pointing arrow.
• Placing the image at the mirror surface instead of at the equal distance behind it.

Things to Be Careful About

• Lateral inversion vs top-bottom inversion: A plane mirror swaps left and right, but not top and bottom. For an asymmetric object (e.g., an arrow pointing right), the image would point left. For a symmetric upward arrow, the image simply points up.
• Always check both the orientation and the distance of the image relative to the mirror.

Walkthrough

1. Identify the angles of incidence and refraction. In optics, angles are always measured from the normal (the dashed line perpendicular to the boundary), not from the boundary surface itself. The diagram gives the angle between the incident ray and the boundary as $42.0^\circ$, so the angle of incidence is $i = 90^\circ - 42.0^\circ = 48.0^\circ$. Similarly, the angle between the refracted ray and the boundary is $49.5^\circ$, so the angle of refraction is $r = 90^\circ - 49.5^\circ = 40.5^\circ$.

2. Calculate the refractive index. Light is bending towards the normal as it enters medium Q ($40.5^\circ < 48.0^\circ$), which means medium Q is optically denser than air. We can find the refractive index $n$ of medium Q using Snell's law:

1. Calculate the critical angle. The critical angle $c$ is the angle of incidence in the denser medium (medium Q) for which the angle of refraction in the less dense medium (air) is $90^\circ$. The formula is $\sin c = \frac{1}{n}$. Substituting the value of $n$:

This matches option C.

Key Takeaways

• Angles are measured from the normal: Always subtract the given angle (if measured from the surface) from $90^\circ$ to get the angle of incidence or refraction.
• Refractive index and critical angle: The refractive index $n$ can be found from the ratio of sines of angles, and the critical angle is found using $\sin c = 1/n$.
• Direction of bending: Light bending towards the normal indicates entering a denser medium, which is a prerequisite for total internal reflection and a critical angle to exist when light tries to exit back into the less dense medium.

Common Mistakes

• Using angles from the boundary directly: A common error is to use $42.0^\circ$ and $49.5^\circ$ directly in Snell's law. This gives an incorrect refractive index and a wrong critical angle.
• Wrong critical angle formula: Some students might try to use $\sin c = n$ or forget to take the reciprocal. Remember $\sin c = 1/n$ only when going from the medium to air (or vacuum).
• Confusing the direction of light: The critical angle is only defined for light travelling from a denser medium to a less dense medium. Here, light must travel from Q to air. The calculation $\sin c = \sin r / \sin i$ correctly captures this because $n = \sin i / \sin r$ for air to Q, so $1/n = \sin r / \sin i$.

Things to Be Careful About

• Read the diagram carefully: The angles $42.0^\circ$ and $49.5^\circ$ are marked between the rays and the horizontal boundary line, not the vertical normal. Always check where the arc is drawn.
• Significant figures: The given angles are to 1 decimal place. The final answer should be given to 1 decimal place as well ($60.9^\circ$), matching the options.
• Unit consistency: Angles are in degrees, ensure your calculator is in degree mode when calculating sines and inverse sines.

Walkthrough

This question tests whether you can pick out the two waves that belong to the electromagnetic spectrum. The electromagnetic spectrum is the full range of electromagnetic waves, from radio waves to gamma rays. The main regions, in order of increasing frequency, are:

radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays.

Now check each option:

• A says light and sound. Light is electromagnetic, but sound is a mechanical wave that needs a medium such as air to travel through.
• B says ultrasound and ultraviolet. Ultraviolet is electromagnetic, but ultrasound is a sound wave, not an electromagnetic wave.
• C says water waves and infrared. Infrared is electromagnetic, but water waves are mechanical waves moving on the surface of water.
• D says X-rays and microwaves. Both are part of the electromagnetic spectrum, so this is the correct answer.

Key Takeaways

• Memorise the main regions of the electromagnetic spectrum: radio, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays.
• Electromagnetic waves can travel through empty space.
• Sound waves and water waves are mechanical waves, so they are not part of the electromagnetic spectrum.

Common Mistakes

• Confusing ultrasound with ultraviolet. Ultrasound is a sound wave; ultraviolet is an electromagnetic wave.
• Assuming that any wave is electromagnetic. Sound and water waves are mechanical and need a medium to travel through.

Things to Be Careful About

• Remember that the electromagnetic spectrum includes only electromagnetic waves, not sound waves or water waves.
• When a question asks for two components, check both parts of the option before choosing it.

Walkthrough

Induced magnetism occurs when an unmagnetised magnetic material, such as an iron nail, is placed in an external magnetic field. The field causes the domains in the iron to align, turning the nail into a temporary magnet.

• Diagram A: The iron nail is placed directly above a permanent bar magnet. The permanent magnet produces a magnetic field, which acts on the nail and induces magnetism in it.
• Diagram B: The iron nail is placed above an unmagnetised iron bar. Because the bar is not magnetised, there is no external magnetic field to induce magnetism in the nail.
• Diagram C: The iron nail is above a solenoid, but the switch is open. With the switch open, no current flows through the coil, so the solenoid produces no magnetic field. The nail remains unmagnetised.
• Diagram D: The iron nail is inside a solenoid, but again the switch is open. No current means no magnetic field, so no induced magnetism occurs.

Therefore, the nail is only an induced magnet in diagram A.

Key Takeaways

• Induced magnetism requires an external magnetic field to align the magnetic domains in a magnetic material.
• Permanent magnets and current-carrying coils (with closed switches) provide the necessary magnetic field; unmagnetised materials and open circuits do not.

Common Mistakes

• Assuming that being near any metal object (like the iron bar in B) will cause induced magnetism, forgetting that the external object must itself be magnetic.
• Forgetting to check the state of the switch in the solenoid diagrams (C and D); an open switch means no current and no magnetic field.

Things to Be Careful About

• Remember that induced magnets are temporary: they lose their magnetism once removed from the external magnetic field.
• Always check both the nature of the external object (permanent magnet vs. unmagnetised material) and the circuit state (switch closed vs. open) when evaluating magnetic field sources.

Walkthrough

This question asks you to choose the row that correctly pairs a magnetic metal with a non-magnetic metal. A magnetic metal is one that is attracted to a magnet, such as iron, nickel and cobalt. Steel contains mostly iron, so steel is also magnetic. Copper and aluminium are not attracted to a magnet, so they are non-magnetic metals.

Look at each option:

• A: aluminium is not magnetic, so this row is already wrong.
• B: copper is not magnetic and steel is magnetic, so the two labels are swapped.
• C: iron is magnetic and copper is non-magnetic, so this row is correct.
• D: steel and iron are both magnetic, so the second label is wrong.

Therefore the correct answer is C.

Key Takeaways

• Common magnetic metals include iron, nickel, cobalt and steel.
• Common non-magnetic metals include copper, aluminium, gold and silver.
• A magnetic material is attracted to a magnet; a non-magnetic material is not.

Common Mistakes

• Thinking that all metals are magnetic. Copper and aluminium are metals but they are not attracted to magnets.
• Thinking that steel is non-magnetic. Steel contains iron, so it is magnetic.
• Confusing iron with copper when recalling examples of magnetic metals.

Things to Be Careful About

• The question asks for a magnetic metal and a non-magnetic metal, so both parts of the row must be correct.
• Steel is an alloy that behaves as a magnetic material because of the iron it contains.
• Aluminium is a metal, but it is not magnetic, so it cannot be the magnetic metal in any correct row.

Walkthrough

1. Direction of the magnetic field: The diagram shows magnetic field lines directed from U to V. In the plane of the page, this is downwards.
2. Alignment of the magnet: A freely suspended magnet will rotate until its north pole points in the direction of the magnetic field lines. Currently, the N pole is on the left and the S pole is on the right. For the N pole to point downwards (towards V), it must move to the right and down, while the S pole must move to the left and up. Viewed from above, this is an anticlockwise rotation of 90°.
3. Translational motion: A common misconception is that the magnet will be pulled towards one end of the field. However, in a uniform magnetic field, the force on the N pole (downwards) is exactly balanced by the equal and opposite force on the S pole (upwards). The resultant force is zero, so there is no net movement towards U or V. The only effect is the couple that turns the magnet.
4. Conclusion: The magnet turns 90° anticlockwise and stops. This matches option B.

Key Takeaways

• A freely suspended magnet aligns its north pole with the external magnetic field.
• A uniform magnetic field exerts equal and opposite forces on the poles of a magnet, producing a turning effect (couple) but no resultant translational force.
• When determining rotation, always use the viewpoint specified (here, 'viewed from above').

Common Mistakes

• Clockwise vs anticlockwise: Choosing option A or C by misjudging the direction of rotation. Remember that the N pole must follow the field lines (downwards in this diagram), which requires an anticlockwise turn from the initial left-right orientation.
• Net translational force: Choosing option C or D by assuming the magnet will be attracted towards one side. In a uniform field, the forces on the two poles cancel out completely; there is no net pull towards U or V.

Things to Be Careful About

• Viewpoint: The question specifies 'viewed from above'. This means clockwise and anticlockwise are judged exactly as they appear on the printed diagram. If the question had said 'viewed from below', the rotation direction would be reversed.
• Uniform vs non-uniform field: If the field were non-uniform (e.g., near a single pole), there would be a resultant force and the magnet would translate as well as rotate. The word 'uniform' is the key to knowing the translation is zero.

Walkthrough

The question tests the model of charging by friction. When two different materials are rubbed together, electrons are transferred from one material to the other. The material that gains electrons becomes negatively charged, while the material that loses electrons becomes positively charged. The key is that only electrons move; protons are tightly bound in the nucleus and do not move during rubbing.

• Option A is incorrect because it says the rod becomes negatively charged, but the rod becomes positively charged.
• Option B is incorrect for the same reason: it says the rod becomes negatively charged.
• Option C is incorrect because it says the rod gains protons, which does not happen during rubbing.
• Option D correctly identifies that the rod loses electrons and becomes positively charged.

Key Takeaways

• Charging by friction involves the transfer of electrons only.
• Loss of electrons results in a positive charge; gain of electrons results in a negative charge.
• Protons do not move during charging by friction.

Common Mistakes

• Choosing B might happen if a student thinks the rod becomes negatively charged, but the rod becomes positively charged.
• Choosing C might happen if a student forgets that protons do not move during rubbing.

Things to Be Careful About

• Remember that the rod becomes positively charged, so it must have lost electrons.
• The direction of electron transfer determines the sign of the charge on each object.

Walkthrough

1. Identify the circuit configuration: The diagram shows four resistors, each of resistance $R$, connected end-to-end in a single loop with a cell of e.m.f. $E$. This is a series circuit.
2. Calculate total resistance: For resistors in series, the total resistance is the sum of the individual resistances. $R_{total} = R + R + R + R = 4R$.
3. Calculate the circuit current: Apply Ohm's law using the total e.m.f. and total resistance: $I = \frac{E}{R_{total}} = \frac{E}{4R}$.
4. Determine current in each resistor: A fundamental rule of series circuits is that the current is the same at every point. Charge is not 'used up' as it flows through the components. Thus, the current through each individual resistor is also $\frac{E}{4R}$.
5. Select the correct option: Option A correctly lists $\frac{E}{4R}$ for all four resistors from left to right.

Key Takeaways

• Series resistance: Resistors in series add together ($R_{total} = R_1 + R_2 + \dots$).
• Current in series: The current is identical everywhere in a series circuit. It does not decrease as it passes through components.
• Ohm's law: The total current in a circuit is given by $I = \frac{V}{R_{total}}$.

Common Mistakes

• Current is 'used up': A common misconception is that current decreases after passing through each resistor, leading to choices like B or C. Current is a rate of flow of charge; charge is conserved, so the current must be the same throughout the loop.
• Ignoring resistors: Forgetting to add all four resistances and simply using $I = \frac{E}{R}$ leads to option D.

Things to Be Careful About

• Current vs. potential difference: While the current is the same everywhere in this series circuit, the potential difference (p.d.) is shared. The p.d. across each individual resistor would be $\frac{E}{4}$. Do not confuse the current formula with the p.d. divider rule.
• Series vs. parallel: Always check the circuit diagram. If these were in parallel, the total resistance would be $\frac{R}{4}$ and the current through each would be $\frac{E}{R}$, but the diagram clearly shows a series connection.