Physics 5054/12 — May/June 2024

Walkthrough

A micrometer is a measuring instrument with a calibrated screw or spindle. It is used to measure very small lengths accurately, for example the diameter of a thin wire or the thickness of a small object. It is not used to measure area, current or mass, so the correct option is C.

Key Takeaways

• A micrometer is a length-measuring instrument.
• It is useful for measuring small distances to a high degree of precision.
• Knowing the correct instrument for each physical quantity is a basic measurement skill.

Common Mistakes

• Choosing A (area): area is measured using a ruler or measuring tape and then calculated, not directly with a micrometer.
• Choosing B (current): current is measured with an ammeter.
• Choosing D (mass): mass is measured with a balance or scale.

Things to Be Careful About

Do not confuse the instrument called a micrometer with the unit called a micrometre. The unit micrometre is one millionth of a metre, but the micrometer is the device used to measure small lengths.

Walkthrough

1. Determine the direction of motion: Look at Fig. 1. The people are positioned on the left side of the boat, pulling on the ropes. This means the resultant of the tension forces pulls the boat to the left, so the boat is moving to the left along the canal.
2. Analyse the forces using Newton's first law: The question states the boat moves at a constant velocity. According to Newton's first law, if an object moves at constant velocity, the resultant (net) force acting on it must be zero. This means all forces acting on the boat must be balanced.
3. Identify the driving force and drag: The driving force is the horizontal resultant of the two tension forces in the ropes. Since the people are pulling to the left, this resultant force points to the left. To balance this and make the resultant force zero, the drag force (water resistance) acting on the boat must be equal in magnitude and opposite in direction to the resultant tension. Therefore, the drag force must point to the right.
4. Match with the options: Look at the options in Fig. 3 (Options). Vector A points nearly horizontally to the right. This matches our deduction for the drag force.

Key Takeaways

• Constant velocity implies zero resultant force: Whenever a question states an object is moving at a constant velocity (or is stationary), you can immediately conclude that the resultant force is zero. All forward forces are balanced by backward forces.
• Vector diagrams for resultants: When adding vectors, placing them head-to-tail allows you to find the resultant by drawing a line from the tail of the first vector to the head of the last. The sideways (perpendicular) components of the two rope tensions cancel each other out, leaving a horizontal resultant along the canal.
• Drag opposes motion: Drag (or friction/resistance) always acts in the direction opposite to the motion of the object.

Common Mistakes

• Choosing B (drag to the left): A candidate might correctly identify that drag opposes motion but misread the direction of motion from Fig. 1, assuming the boat is moving to the right because the word "boat" is on the right side of the diagram. The people are clearly on the left, pulling leftwards.
• Choosing C or D (drag at an angle): A candidate might forget that the boat is constrained to move along the canal. The vertical components of the two rope tensions cancel each other out (one pulls up-left, the other down-left), so the resultant driving force is purely horizontal. Consequently, the balancing drag force must also be purely horizontal.
• Confusing tension with drag: Selecting a vector that represents the tension forces rather than the reaction drag force.

Things to Be Careful About

• Reading diagrams carefully: Always look at the labels and the position of the actors (the people) to determine the direction of motion. In Fig. 1, the "canal" label is on the left and the people are on the left, indicating leftward motion.
• Constant velocity vs. acceleration: If the boat were accelerating, the resultant force would be in the direction of motion (left), and the drag would be smaller than the resultant tension. Because the velocity is constant, they are exactly equal and opposite.
• Vector diagram interpretation: In Fig. 2, the two tension vectors are shown. Even if the exact drawing is slightly abstract, the physical constraint that the boat moves straight along the canal means the vertical components must cancel, leaving a horizontal resultant.

Walkthrough

The question asks for the section with the largest acceleration on a speed-time graph. We need to recall the relationship between the graph's features and physical quantities.

1. Identify the physical quantity represented by the gradient: On a speed-time graph, the vertical axis is speed ($v$) and the horizontal axis is time ($t$). The gradient (slope) is $\frac{\text{change in speed}}{\text{change in time}} = \frac{\Delta v}{\Delta t}$. This is the definition of acceleration.
2. Analyze each section:
   • Section A: The graph rises steeply. A steep positive gradient means a large positive acceleration.
   • Section B: The graph rises gently. A gentle positive gradient means a small positive acceleration.
   • Section C: The graph is horizontal. The gradient is zero, so the acceleration is zero (the car is moving at constant speed).
   • Section D: The graph falls. The gradient is negative, which represents deceleration (slowing down).
3. Compare: The question asks for the largest acceleration. This corresponds to the steepest positive slope. Section A is clearly steeper than section B. Section D represents a reduction in speed, not an increase (unless we are talking about magnitude of deceleration, but usually 'largest acceleration' in this context implies the greatest rate of speeding up, and visually A is steeper than D anyway).

Therefore, section A shows the largest acceleration.

Key Takeaways

• On a speed-time graph, the gradient represents acceleration.
• A steeper slope means a larger acceleration.
• A horizontal line means zero acceleration (constant speed).
• A falling slope means deceleration (negative acceleration).

Common Mistakes

• Confusing speed-time with distance-time: On a distance-time graph, the gradient is speed. On a speed-time graph, the gradient is acceleration.
• Confusing speed with acceleration: Section C has the highest speed, but the acceleration is zero because the speed is not changing.
• Ignoring the sign: Section D has a negative gradient (deceleration). While it has a large magnitude of change, 'acceleration' usually refers to speeding up in positive direction in simple contexts, or simply the steepest positive slope. Even if magnitude is considered, A is steeper than D.

Things to Be Careful About

• Always check the axis labels. Y-axis is 'speed', X-axis is 'time'.
• 'Largest acceleration' means the steepest positive slope. If the question asked for the largest deceleration, you would look for the steepest negative slope.
• Do not assume the section with the highest speed has the highest acceleration.

Walkthrough

The question asks which quantity can be calculated using only distance and time.

1. Average speed is a scalar quantity defined as total distance divided by total time. Since both distance and time are measured, average speed can be calculated directly.
2. Average velocity is a vector quantity defined as displacement divided by time. Displacement is the straight-line distance from the starting point to the finishing point, in a specific direction. The diagram shows a curved path from X to Y, meaning the distance walked is longer than the straight-line displacement. Because the man only measures the distance along the path, he does not know the displacement, so average velocity cannot be calculated.
3. Acceleration is the rate of change of velocity. Calculating it requires knowing the initial and final velocities or the change in velocity over time, which is not provided.
4. Power is the rate of doing work or transferring energy. Calculating it requires knowing the work done (force times displacement) or energy used, neither of which is provided.

Therefore, only average speed can be calculated from the given data.

Key Takeaways

• Distance is a scalar (total path length); displacement is a vector (straight-line change in position with direction).
• Speed is a scalar (distance / time); velocity is a vector (displacement / time).
• You can only calculate a quantity if you have all the necessary variables. Missing a vector component (like direction or straight-line displacement) means you cannot calculate the corresponding vector quantity.

Common Mistakes

• Choosing average velocity because students confuse distance with displacement. Remember that velocity requires displacement, which is the shortest straight-line distance between two points with a direction. The curved path in the diagram is a hint that distance $\neq$ displacement.
• Choosing acceleration without realising it requires a change in velocity, not just distance and time.

Things to Be Careful About

• Always check whether the quantity asked for is a scalar or a vector. Distance and time are both scalars, so their ratio (average speed) is a scalar. Displacement is a vector, so its ratio with time (average velocity) is a vector.
• Read diagrams carefully: a curved or winding path means the distance travelled is strictly greater than the magnitude of the displacement.

Walkthrough

The question asks which property determines an object's resistance to a change from its state of rest or motion. This resistance is called inertia. Inertia is a property of mass: a more massive object is harder to start moving, and once moving, harder to stop or change direction. For example, it is much harder to push a loaded lorry than an empty shopping trolley because the lorry has more mass and therefore more inertia. Shape, surface area and volume do not affect inertia.

Key Takeaways

• Inertia is the resistance of an object to a change in its state of rest or motion.
• Mass is the measure of inertia.
• The greater the mass, the greater the inertia, so a larger force is needed to produce the same acceleration.

Common Mistakes

• Choosing volume or surface area because larger objects seem harder to move. In physics, it is mass, not size or shape, that determines inertia.
• Confusing inertia with weight. Weight depends on gravity, but inertia depends only on mass.

Things to Be Careful About

• The word "resistance" here refers to inertia, not to friction or air resistance.
• Even in space where weight is zero, an object still has mass and therefore still has inertia.

Walkthrough

1. Identify the constant quantity: Mass is a measure of the amount of matter in an object and does not change when the object moves from the Earth to the Moon. Therefore, the mass $m$ is $210$ kg on the Moon as well.
2. Calculate weight on Earth: Weight is the force of gravity acting on a mass, given by $W = mg$. Using the standard value $g = 9.8$ N/kg for Earth: $$W_{\text{Earth}} = 210 \text{ kg} \times 9.8 \text{ N/kg} = 2058 \text{ N}$$
3. Calculate weight on the Moon: The problem states the gravitational field strength on the Moon is $\frac{1}{6}$ of that on Earth. Since weight is directly proportional to gravitational field strength ($W = mg$), the weight on the Moon is $\frac{1}{6}$ of the weight on Earth: $$W_{\text{Moon}} = \frac{1}{6} \times 2058 \text{ N} = 343 \text{ N}$$
4. Match with options: $343$ N rounds to $340$ N (to 2 significant figures), which matches option D.

Key Takeaways

• Mass vs. Weight: Mass is constant everywhere; weight changes depending on the local gravitational field strength.
• Weight formula: $W = mg$, where $m$ is mass in kg and $g$ is gravitational field strength in N/kg.
• Proportionality: If $g$ changes by a factor, $W$ changes by the same factor.

Common Mistakes

• Choosing B (35 N): This is the result of dividing the mass by 6 ($210 / 6 = 35$) but forgetting to multiply by the gravitational field strength $g$. The candidate calculated $m/6$ instead of $mg/6$.
• Choosing C (210 N): This is simply the numerical value of the mass, confusing mass (kg) with weight (N), or assuming weight equals mass numerically.
• Choosing A (zero): Assuming that because there is no atmosphere or it's a "moon", there is no gravity. The Moon has gravity, just less than Earth's.

Things to Be Careful About

• Units: Weight is a force measured in Newtons (N), not kilograms (kg). Always ensure the final answer has units of force.
• Value of g: Cambridge O Level Physics typically uses $g = 9.8$ N/kg or $g = 10$ N/kg. Here, using $g = 9.8$ N/kg gives $343$ N, which rounds to the option $340$ N. Using $g = 10$ N/kg gives $350$ N, which is not an option, confirming $9.8$ N/kg is the intended value.
• Significant figures: The options are given to 2 significant figures ($35$, $210$, $340$), so the final calculated value should be rounded appropriately.

Walkthrough

The question gives a forward driving force of $800\text{ N}$ and a backward resistive force $R$. The motion of the car depends on the resultant force, which is the difference between the forward and backward forces: $F_{\text{resultant}} = 800 - R$.

• When $R < 800\text{ N}$, the resultant force is in the forward direction, causing the car to accelerate (Newton's second law).
• When $R = 800\text{ N}$, the forces are balanced, the resultant force is zero, and the car moves at a constant velocity (Newton's first law).
• When $R > 800\text{ N}$, the resultant force is in the backward direction, causing the car to decelerate.

We evaluate each row:

• Row A: $R = 500\text{ N}$. Since $500 < 800$, the resultant force is $300\text{ N}$ forward, so the car accelerates. The row claims deceleration, so A is wrong.
• Row B: $R = 800\text{ N}$. The forces are balanced, resultant force is zero, so the car has constant velocity. The row claims acceleration, so B is wrong.
• Row C: $R = 900\text{ N}$. Since $900 > 800$, the resultant force is $100\text{ N}$ backward, so the car decelerates. The row claims deceleration, so C is correct.
• Row D: $R = 1000\text{ N}$. Since $1000 > 800$, the resultant force is $200\text{ N}$ backward, so the car decelerates. The row claims acceleration, so D is wrong.

Key Takeaways

• The resultant force determines the motion of an object: a forward resultant force means acceleration, a backward resultant force means deceleration, and a zero resultant force means constant velocity.
• Newton's first and second laws apply directly to situations with opposing forces like driving and resistive forces.

Common Mistakes

• Confusing the direction of the resultant force: forgetting that resistive forces oppose motion, so if $R > 800\text{ N}$, the resultant is backward, not forward.
• Thinking that any forward force always causes acceleration, without considering the resistive force.
• Assuming that if $R = 800\text{ N}$, the car must be accelerating because the engine is still providing force, forgetting that balanced forces mean constant velocity.

Things to Be Careful About

• Ensure the direction of the resultant force is correctly assigned: forward is positive (acceleration in the direction of motion), backward is negative (deceleration).
• Remember that "constant velocity" is not the same as "acceleration" or "deceleration"; it is the state when forces are balanced.
• Read the table carefully to match the calculated motion with the description given in the row.

Walkthrough

The four forces listed are:

• P: force on the Moon due to the Earth
• Q: force on the Earth due to the Sun
• R: force on the Earth due to the Moon
• S: force on the Moon due to the Sun

Newton's third law says that if one object pulls on another, the second object pulls back on the first with an equal and opposite force. So the two forces in a pair must act between the same two objects.

P and R both involve the Earth and the Moon. P is the Earth pulling the Moon, and R is the Moon pulling the Earth. These are equal in size, opposite in direction, and act on different objects, so they form a Newton's third law pair.

The other options pair forces that involve different pairs of objects, so they cannot be a Newton's third law pair.

Key Takeaways

• A Newton's third law pair always involves the same two objects.
• The two forces act on different objects, so they do not cancel each other out.
• Gravitational forces always come in equal and opposite pairs.

Common Mistakes

• Choosing Q and S because both involve the Sun. Q is the force on the Earth due to the Sun, but S is the force on the Moon due to the Sun. They are not a pair because they do not involve the same two objects.
• Choosing P and Q because both involve Earth and Moon, but Q involves the Sun, not the Moon.
• Thinking the pair must be the same type of force. They are both gravitational, but that is not enough — the two objects must be the same pair.

Things to Be Careful About

• Read the phrases carefully: "force on X due to Y" identifies both the object feeling the force and the object causing it.
• The third-law partner of "force on X due to Y" is "force on Y due to X".
• Equal and opposite forces do not cancel because they act on different bodies.

Walkthrough

The box is moving up the slope from X to Y. Friction is a resistive force that always acts in the direction opposite to the relative motion between two surfaces in contact. Because the box is moving up the slope, the friction force must act down the slope, towards X. Looking at the diagram, arrow C points down the slope towards X, so C is the friction force.

The other arrows represent different forces:

• Arrow A points up the slope towards Y; this is the applied force pulling or pushing the box up.
• Arrow B points vertically downwards; this is the weight of the box (the force due to gravity).
• Arrow D points perpendicular to the slope, away from it; this is the normal reaction force exerted by the slope on the box.

Key Takeaways

• Friction always opposes the direction of motion. If an object moves up a slope, friction acts down the slope.
• Weight always acts vertically downwards towards the centre of the Earth, regardless of the orientation of the surface.
• The normal reaction force always acts perpendicular to the surface in contact, pushing away from it.

Common Mistakes

• Choosing arrow A as friction: confusing the direction of motion with the direction of friction. Friction opposes motion, so it must point down the slope.
• Choosing arrow B as friction: confusing weight with friction. Weight acts vertically downwards, not along the slope.
• Choosing arrow D as friction: confusing normal reaction with friction. Normal reaction is perpendicular to the surface, while friction is parallel to it.

Things to Be Careful About

• Always determine the direction of motion first before deciding the direction of friction. Friction opposes motion, not necessarily the applied force (though in this simple case they are opposite).
• Remember that weight is always vertical, not perpendicular to the slope. The component of weight acting down the slope is not explicitly drawn here as a separate arrow, but friction is a separate force acting parallel to the surface.

Walkthrough

The question asks which thread is the most difficult to stretch. In terms of a load-extension graph, 'difficult to stretch' means that a large load is required to produce a small extension. Equivalently, for any given load on the horizontal axis, the extension on the vertical axis is the smallest.

Looking at the graph:

• Curve A is a straight line with a steep gradient. A small load produces a large extension, so it is easy to stretch.
• Curve B starts with a steep gradient and then curves over, but still shows a large extension for moderate loads.
• Curve C starts with a shallow gradient but curves upwards, meaning it becomes easier to stretch as the load increases. Over the full range, it does not have the smallest extension.
• Curve D has a very shallow gradient throughout the range shown. For any given load, the extension is the smallest of all four curves. This means it is the stiffest and most difficult to stretch.

Thus, the correct option is D.

Key Takeaways

• On an extension vs. load graph, a shallow gradient indicates a stiff material (hard to stretch, high spring constant), while a steep gradient indicates a flexible material (easy to stretch, low spring constant).
• Always read the axes carefully: extension is on the y-axis and load is on the x-axis here, so the gradient is extension per unit load (the inverse of stiffness).

Common Mistakes

• Confusing the axes: if a candidate reads load on the y-axis and extension on the x-axis, they would choose the curve with the steepest gradient. Always check which quantity is on which axis.
• Choosing C because it has a shallow gradient at the start: the question asks over the range of loads shown, and D has the smallest extension across the entire range.
• Thinking 'most difficult to stretch' means 'breaks first' or 'has the largest extension': 'difficult to stretch' refers to stiffness, not ultimate strength.

Things to Be Careful About

• The axes are extension (y) and load (x). The gradient of this graph is $\frac{\Delta \text{extension}}{\Delta \text{load}}$, which is the inverse of the spring constant $k$ (since $k = \frac{\text{load}}{\text{extension}}$). A smaller gradient means a larger $k$, hence a stiffer spring.
• Pay attention to the wording 'over the range of loads shown'. Curve C is shallow at first, but D is shallow throughout the whole range.

Walkthrough

The satellite is moving in a circular orbit around the planet. For any object to move in a circular path at constant speed, there must be a resultant force acting on it directed towards the centre of the circle. This is the centripetal force. In the case of a satellite orbiting a planet, this centripetal force is provided by the gravitational attraction between the planet and the satellite. Gravity is an attractive force, so it pulls the satellite towards the centre of the planet. Looking at the diagram, the satellite is at the top right and the planet is at the centre. The direction from the satellite towards the centre of the planet is downwards and to the left. Arrow D points in this direction. Therefore, the force on the satellite is in the direction D.

Key Takeaways

• An object moving in a circular path requires a resultant force directed towards the centre of the circle (centripetal force).
• For a satellite in orbit, this force is the gravitational pull of the planet it is orbiting, which is always attractive and acts along the line connecting the two bodies towards the centre of the planet.

Common Mistakes

• Confusing the direction of the force with the direction of motion (velocity). The velocity is tangential to the circle (in the direction of the arrow labelled 'direction of movement'), but the force is towards the centre.
• Choosing an arrow pointing away from the centre. Centrifugal force is not a real force in this inertial frame; the only force acting is the inward gravitational pull.

Things to Be Careful About

• Remember that the force causing circular motion is always directed towards the centre, not along the path of motion.
• Gravity is always an attractive force, so it can only pull objects together, never push them apart.

Walkthrough

The question asks for the horizontal distance $d$ between the 2300 N load and the centre of gravity P such that there is no net moment about P. This means the crane jib is in rotational equilibrium about P, so we apply the principle of moments: the total clockwise moment about P equals the total anticlockwise moment about P.

The anticlockwise moment is produced by the 5800 N counterweight at a perpendicular distance of 2.3 m from P:

The clockwise moment is produced by the 2300 N load at an unknown distance $d$ from P:

Equating the two moments:

Solving for $d$:

This matches option C.

Key Takeaways

• The principle of moments states that for an object in rotational equilibrium about a pivot, the sum of the clockwise moments equals the sum of the anticlockwise moments.
• The moment of a force is calculated as $\text{moment} = \text{force} \times \text{perpendicular distance from the pivot}$.

Common Mistakes

• Forgetting to equate clockwise and anticlockwise moments and instead adding them together.
• Using the wrong distance or force in the moment calculation (e.g., measuring distance from the tower instead of from the pivot P).
• Calculation errors when dividing 13 340 by 2300.

Things to Be Careful About

• Ensure all distances are measured from the pivot (point P), not from the ends of the jib or the tower base.
• The unit of moment is newton metres (N m), but since we are equating moments on both sides of the equation, the units cancel out when solving for distance, leaving metres (m) as the final unit for $d$.

Walkthrough

Momentum is the product of mass and velocity: $p = m v$. The question asks for the change in momentum during the acceleration, so only the change in velocity matters: the car speeds up from $25\ \text{m/s}$ to $35\ \text{m/s}$, a change of $10\ \text{m/s}$. Multiply by the mass of the car:

$\Delta p = m\Delta v = 750 \times 10 = 7500\ \text{kg m/s}$

The distance travelled (400 m before and after the acceleration) is not needed because momentum depends only on mass and velocity, not on distance. The correct option is A.

Key Takeaways

• Momentum is $p = mv$, and its unit is $\text{kg m/s}$.
• The change in momentum is $\Delta p = m\Delta v$.
• Extra information in a question can be a distractor; identify which quantities actually matter.

Common Mistakes

• Using the final speed only: $750 \times 35 = 26\,250\ \text{kg m/s}$, which is not an option.
• Adding the speeds: $25 + 35 = 60$, then $750 \times 60 = 45\,000\ \text{kg m/s}$ (option C).
• Using the distance 400 m as if it were a speed.
• Forgetting the unit $\text{kg m/s}$.

Things to Be Careful About

• The change in momentum uses the change in velocity, not the final velocity.
• Momentum is a vector, but here the car does not change direction, so the arithmetic difference is correct.
• The unit must be included: $\text{kg m/s}$.
• The 400 m values are repeated distractors; do not let them affect the calculation.

Walkthrough

The ball starts from rest, so its initial velocity is $u = 0$. Since air resistance is negligible, the only acceleration is the acceleration of free fall, $g = 10\ \text{m/s}^2$. The final velocity is given as $v = 14\ \text{m/s}$, and the height of the building is the vertical distance the ball falls.

Because the acceleration is constant and the time is not given, use the equation that connects velocity, acceleration and displacement:

Here $s$ is the displacement, which is the height of the building. Substituting $u = 0$, $v = 14\ \text{m/s}$ and $a = 10\ \text{m/s}^2$:

This rounds to $10\ \text{m}$, so the correct option is B.

Key Takeaways

• Free fall without air resistance is a constant-acceleration motion with acceleration $g$.
• When the time is not given, use $v^2 = u^2 + 2as$.
• The height of the building is the displacement of the ball, not its final velocity.

Common Mistakes

• Forgetting to square the velocity when substituting into $v^2 = u^2 + 2as$.
• Using $s = \frac{v^2}{g}$ instead of $\frac{v^2}{2g}$; this gives about $20\ \text{m}$, which is option C.
• Using $s = \frac{2v^2}{g}$; this gives about $40\ \text{m}$, which is option D.
• Confusing the final velocity with the average velocity.

Things to Be Careful About

• Use $g = 10\ \text{m/s}^2$ unless the question states otherwise. Here both $g = 10$ and $g = 9.8$ lead to option B.
• Since the ball is dropped, $u = 0$; do not treat the final velocity as the initial velocity.
• Always include the unit, $\text{m}$, when giving the height.
• The phrase "air resistance is negligible" is important: it tells you the acceleration is constant and equal to $g$.

Walkthrough

The question asks which energy source is available constantly over a 24-hour period. This is about availability, not about whether the source is renewable.

• Natural gas is a fuel that can be stored and burned at any time of day or night, so it is available whenever it is needed.
• Solar cells only generate electricity when the Sun is shining, so they are not available at night or in cloudy weather.
• Tidal power only works when the tide is moving in or out, which happens at certain times of the day, not constantly.
• Wind power only works when the wind is blowing, and the wind is not always present.

Therefore the correct answer is A, natural gas.

Key Takeaways

• Some energy sources are intermittent (solar, wind, tidal) because they depend on conditions that are not always present.
• Other sources, such as natural gas, are continuously available because they can be stored and used when needed.
• The question is about availability over a 24-hour period, not about whether the source is renewable.

Common Mistakes

• Choosing solar cells because they are renewable, without noticing that they only work during daylight.
• Choosing wind because it is clean, without noticing that wind is not constant.
• Choosing tidal because it is predictable, but tidal power is not available all the time.
• Thinking that "natural gas" is wrong because it is a fossil fuel, but the question does not ask about renewable energy.

Things to Be Careful About

• Read the question carefully: it asks about availability over 24 hours, not about whether the source is renewable or environmentally friendly.
• Natural gas is a non-renewable source, but it can still be available constantly while the supply lasts.
• Solar, wind, and tidal are renewable but intermittent, so they do not satisfy the condition of being available constantly.

Walkthrough

A 15 W lamp transfers 15 J of energy every second. It is turned on for 30 minutes, so first convert this to seconds:

$30 \times 60 = 1800\ \text{s}$

The total energy supplied to the lamp is therefore

$\text{energy in} = 15 \times 1800 = 27000\ \text{J}$

Of this total, 7000 J is wasted. So the useful energy output is

$27000 - 7000 = 20000\ \text{J}$

Efficiency is the fraction of the energy input that is converted into useful output:

$\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}} = \frac{20000}{27000} = 0.74$

As a percentage, this is 74%. The correct option is D.

Key Takeaways

• Power tells you the rate of energy use: 15 W means 15 J of energy each second.
• Total energy input is found from $\text{power} \times \text{time}$.
• Efficiency is always useful output divided by total input, never wasted energy divided by input.
• Efficiency has no unit because it is a ratio.

Common Mistakes

• Using 30 minutes instead of 1800 seconds. This would give $15 \times 30 = 450\ \text{J}$, which is wrong.
• Dividing the wasted energy by the total input: $7000/27000 = 0.26$, which is option A.
• Dividing the wasted energy by the useful energy: $7000/20000 = 0.35$, which is option B.
• Forgetting that the useful energy must be found by subtracting the wasted energy from the total energy.

Things to Be Careful About

• Always convert minutes to seconds before using $E = Pt$.
• Use useful output, not wasted energy, in the numerator of the efficiency.
• The answer is a decimal between 0 and 1, so no unit is needed.
• In multiple-choice questions, check that your final decimal matches one of the options exactly.

Walkthrough

This question tests whether you know the units of common energy-related quantities.

• Gravitational potential energy is a form of energy, so it is measured in joules.
• Latent heat is the energy needed to change the state of a substance without changing its temperature, so it is also measured in joules.
• Work is the transfer of energy, so it is measured in joules.
• Power is not an amount of energy; it is the rate at which energy is transferred. Its unit is the watt (W), which is equal to one joule per second.

Therefore, the quantity that is not measured in joules is power.

Key Takeaways

• Energy, including gravitational potential energy and latent heat, is measured in joules (J).
• Work is energy transferred, so it is also measured in joules.
• Power is the rate of energy transfer and is measured in watts (W), where 1 W = 1 J/s.

Common Mistakes

• Choosing "work" because it seems different from energy, even though work is a measure of energy transfer and has the unit joule.
• Confusing power with energy. Power tells you how quickly energy is transferred, not how much energy is involved.

Things to Be Careful About

• Remember the definitions: energy and work are measured in joules; power is measured in watts.
• A watt is a joule per second, so power can also be expressed as J/s, but it is not simply a quantity of joules.

Walkthrough

The question asks for the correct description of a liquid. In the kinetic particle model, a liquid has particles that are close together but not held in fixed positions. This means the particles can slide over each other, so the liquid does not have a fixed shape and takes the shape of its container. However, the particles cannot spread far apart, so the liquid keeps a fixed volume. Therefore the correct answer is C: variable shape, fixed volume.

Option A describes a solid, which has both fixed shape and fixed volume. Option B describes something with fixed shape but variable volume, which is not a normal state of matter. Option D describes a gas, which has both variable shape and variable volume.

Key Takeaways

The key idea is that each state of matter has a specific combination of shape and volume. Solids have fixed shape and fixed volume. Liquids have variable shape but fixed volume. Gases have variable shape and variable volume. These properties come from how the particles are arranged and how freely they move.

Common Mistakes

• Choosing A because a liquid seems to have a fixed volume, but A also says fixed shape, which is wrong for a liquid.
• Choosing D because a liquid can flow, but flowing means the shape changes, not the volume.
• Thinking that a liquid has variable volume because it can be poured into different containers. The volume stays the same; only the shape changes.

Things to Be Careful About

• Shape and volume are two separate properties. A liquid can change shape but its volume is fixed under normal conditions.
• The volume of a liquid can change slightly with temperature, but the question is asking about the usual description of a liquid.
• Do not confuse the description of a liquid with that of a gas: both have variable shape, but only a gas has variable volume.

Walkthrough

Gas particles are in constant, random motion. When the dividing wall is removed, the particles of the brown gas move into the right-hand compartment and the particles of the colourless gas move into the left-hand compartment. Over a long time, the random motion causes the two gases to mix completely and uniformly throughout the entire box. Since the brown gas is now spread evenly through the larger volume, its colour is diluted, resulting in a pale brown gas throughout.

Key Takeaways

• Gas particles move randomly and continuously.
• Diffusion is the net movement of particles from a region of higher concentration to lower concentration until a uniform mixture is achieved.
• When two gases mix, the coloured gas will dilute to a pale colour throughout the available volume.

Common Mistakes

• Assuming the heavier brown gas will sink or stay on the left side. Gas particles move randomly in all directions, and diffusion dominates over any minor density effects in a small box over a long time.
• Thinking the gases will remain separate or form distinct clouds. Complete mixing occurs over a long time.

Things to Be Careful About

• The phrase left for a long time ensures that complete diffusion and mixing have occurred.
• The gases do not react, so they simply mix without forming a new substance.

Walkthrough

On a warm day, the air contains water vapour. The cold liquid in the bottle cools the air in contact with the outside of the bottle. When this air is cooled enough, the water vapour changes from a gas into tiny liquid droplets. This change of state is called condensation.

The other options are wrong. Conduction and convection are methods of transferring thermal energy, not changes of state. Evaporation is a liquid changing into a gas, which is the opposite of condensation.

Key Takeaways

Condensation is the change of state from gas to liquid. It happens when a gas is cooled enough, often by touching a cooler surface. Evaporation is the opposite change, from liquid to gas. Conduction and convection are heat transfer processes, not changes of state.

Common Mistakes

• Choosing evaporation because water droplets appear. Evaporation would make water disappear as a gas, not form drops.
• Choosing conduction or convection because the day is warm. These are heat transfer processes, but they do not turn water vapour into liquid.
• Thinking the drops leak from inside the bottle. The drops are on the outside, so they form from water vapour in the air.

Things to Be Careful About

The drops are outside the bottle, so they cannot come from the liquid inside. The cold surface cools the air around it, causing water vapour to condense. Condensation is a change of state, not a method of heat transfer. Also, do not confuse the words "condensation" and "convection" — they sound similar but have different meanings.

Walkthrough

The question asks for the definition of specific heat capacity. Specific heat capacity is a property of a material: it tells us how much energy is needed to raise the temperature of $1\ \text{kg}$ of that material by $1\,^\circ\text{C}$.

The defining equation is

where $E$ is the thermal energy transferred, $m$ is the mass and $\Delta\theta$ is the temperature change. Option C matches this definition exactly.

Option A is just the difference between two temperatures, so it is not an energy per unit mass. Option B describes specific latent heat, which is the energy needed to change state without changing temperature. Option D describes a temperature change, not the energy needed to produce it.

Key Takeaways

• Specific heat capacity is the energy needed to raise the temperature of $1\ \text{kg}$ of a substance by $1\,^\circ\text{C}$.
• It is measured in $\text{J/(kg}\,^\circ\text{C)}$.
• It is different from specific latent heat, which is the energy needed to change state.

Common Mistakes

• Choosing B: B describes specific latent heat, not specific heat capacity.
• Choosing D: a temperature rise alone does not define specific heat capacity, because it does not mention mass or energy.
• Writing the unit as just $\text{J}$: the unit must include mass and temperature change.

Things to Be Careful About

• A temperature change of $1\,^\circ\text{C}$ is the same size as a change of $1\ \text{K}$, so either can be used in the definition.
• Specific heat capacity is per kilogram, so it does not depend on the total mass of the sample.
• Do not confuse specific heat capacity with heat capacity, which is the energy needed to warm a whole object by $1\,^\circ\text{C}$.

Walkthrough

This question tests four separate facts about infrared radiation, and you must judge each statement before picking the option.

Statement A — In a vacuum, infrared travels at the speed of light. All electromagnetic waves — radio, microwave, infrared, visible, ultraviolet, X-rays and gamma rays — are the same kind of wave, and they all travel at $3.0 \times 10^{8}\ \text{m/s}$ in a vacuum. So A is correct.

Statement B — Infrared is a longitudinal wave. All electromagnetic waves are transverse: the oscillations are perpendicular to the direction of travel. Longitudinal waves (like sound) oscillate along the direction of travel. So B is false.

Statement C — Infrared has a higher frequency than ultraviolet. The electromagnetic spectrum is ordered by frequency and wavelength. From lowest frequency to highest: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma. Ultraviolet sits above infrared, so ultraviolet has the higher frequency. C is false.

Statement D — White surfaces are better emitters than black surfaces. The opposite is true: black (dull) surfaces are the best emitters and absorbers of infrared radiation, while white (shiny) surfaces reflect more and emit less. D is false.

Since only statement A is correct, the answer is A.

Key Takeaways

• All electromagnetic waves travel at the speed of light in a vacuum ($3.0 \times 10^{8}\ \text{m/s}$).
• All electromagnetic waves are transverse waves.
• The electromagnetic spectrum in order of increasing frequency: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma.
• Black (dull) surfaces are better emitters and absorbers of infrared radiation than white (shiny) surfaces.

Common Mistakes

• Confusing the order of the spectrum — thinking infrared has a higher frequency than ultraviolet. Remember infrared sits just below visible light and ultraviolet just above it.
• Thinking infrared is longitudinal because sound is longitudinal. All electromagnetic waves, including infrared, are transverse.
• Thinking white surfaces emit infrared better. Black surfaces are the best emitters and absorbers; white surfaces are better reflectors.

Things to Be Careful About

• Frequency and wavelength are inversely related: higher frequency means shorter wavelength. Ultraviolet has higher frequency and shorter wavelength than infrared.
• The speed of light in a vacuum is $3.0 \times 10^{8}\ \text{m/s}$ — recall this exact value.
• "Speed of light" applies to the whole electromagnetic spectrum, not just visible light.
• In the exam, read each option carefully — one wrong fact in an option makes the whole statement false.

Walkthrough

1. Recall what ultrasound means. Ultrasound is sound with a frequency higher than the upper limit of human hearing, about $20\,000\ \text{Hz}$. So I must find the frequency of each sound and see whether any frequency is greater than $20\,000\ \text{Hz}$.

2. Use the wave equation. For any wave, $v = f \lambda$, where:

   • $v$ is the wave speed, here $330\ \text{m/s}$,
   • $f$ is the frequency,
   • $\lambda$ is the wavelength.

   To find frequency, rearrange: $f = \dfrac{v}{\lambda}$.

3. Convert all wavelengths to metres first. The speed is in metres per second, so the wavelength must also be in metres.

   • $250\ \text{cm} = 2.5\ \text{m}$,
   • $25\ \text{cm} = 0.25\ \text{m}$,
   • $2.5\ \text{cm} = 0.025\ \text{m}$.

4. Calculate each frequency.

   • A: $f = 330/2.5 = 132\ \text{Hz}$.
   • B: $f = 330/0.25 = 1320\ \text{Hz}$.
   • C: $f = 330/0.025 = 13\,200\ \text{Hz}$.

5. Compare with the ultrasound threshold. Ultrasound is greater than $20\,000\ \text{Hz}$. Option C is the highest, but even it is only $13,200\ \text{Hz}$, which is still audible. None of the options gives a frequency above $20,000\ \text{Hz}$, so the correct answer is D.

Key Takeaways

• Ultrasound is sound with frequency above the upper limit of human hearing, about $20,000\ \text{Hz}$.
• To decide whether a sound is ultrasound, use $v = f\lambda$ to find its frequency.
• Always convert lengths to metres before substituting into the wave equation when the speed is given in metres per second.
• A long wavelength means a low frequency; a short wavelength means a high-frequency sound.

Common Mistakes

• Forgetting to convert centimetres to metres. For example, using $\lambda = 2.5 \text{cm}$ directly gives a very large frequency and might lead to option C being chosen incorrectly.
• Confusing ultrasonics with "high pitch but still audible". A sound of $13,200\ \text{Hz}$ is high-pitched but still within human hearing.
• Thinking that the correct option must be one of A, B, or C because the question lists three possible wavelengths. The option D exists because none may fit ultrasound.

Things to Be Careful About

• Use the exact threshold $20,000\ \text{Hz}$ for ultrasound.
• Use the wave speed as $330\ \text{m/s}$, as stated in the question.
• Keep units consistent: $1\ \text{cm} = 0.01\ \text{m}$.
• Even though option C gives the highest frequency, it is still below $20,000\ \text{Hz}$, so it is not ultrasound.

Walkthrough

The question asks for the ray diagram showing how light from a distant object forms an image in a normal eye. Light from a distant object arrives as parallel rays.

In a normal eye, the cornea and the crystalline lens act together as a converging lens system. They refract the incoming parallel rays so that they meet at a single point on the retina, which is the light-sensitive back surface of the eye. This produces a sharp, focused image.

Let us evaluate each option:

• Diagram A: The parallel rays are refracted by the lens and converge exactly on the retina. This is the correct diagram for a normal eye.
• Diagram B: The rays pass straight through the lens without bending. A normal eye must refract light to focus it; this diagram is incorrect.
• Diagram C: The rays cross and focus at a point inside the eyeball, in front of the retina. This is the characteristic diagram for myopia (short-sightedness). The image is blurred on the retina.
• Diagram D: The rays have not yet converged when they reach the retina; their focal point is behind the retina. This is the characteristic diagram for hypermetropia (long-sightedness).

Key Takeaways

• A normal eye focuses parallel rays (from a distant object) exactly on the retina.
• Myopia (short-sightedness): image forms in front of the retina. Corrected with a diverging lens.
• Hypermetropia (long-sightedness): image forms behind the retina. Corrected with a converging lens.
• Light from distant objects arrives as parallel rays; light from nearby objects arrives as diverging rays.

Common Mistakes

• Choosing Diagram C or Diagram D and forgetting which condition they represent. Remember: 'myopia' = 'my' (me) = in front; 'hyper' = over/behind. Or simply remember that a normal eye focuses on the retina, myopia before the retina, and hypermetropia behind the retina.
• Selecting Diagram B and thinking the lens has no effect. The eye's lens must refract light to form an image.
• Confusing the direction of the rays: rays from a distant object are parallel, not diverging from a point near the eye.

Things to Be Careful About

• Ensure you read the question carefully: it asks for a normal eye, not a short-sighted or long-sighted one. Diagrams C and D are very common distractors because they are frequently discussed in the context of correcting vision.
• In ray diagrams for the eye, the retina is always the back curved surface of the eyeball. The focal point must lie exactly on this surface for normal vision.

Walkthrough

The question asks for the list of colours in order of increasing wavelength. In the visible spectrum, red light has the longest wavelength (about 700 nm) and violet light has the shortest (about 400 nm). The full order from shortest to longest wavelength is:

violet $\rightarrow$ blue $\rightarrow$ green $\rightarrow$ yellow $\rightarrow$ orange $\rightarrow$ red

Check each option against this order:

• A blue $\rightarrow$ green $\rightarrow$ yellow $\rightarrow$ red — each step moves to a longer wavelength, so this is correct.
• B blue $\rightarrow$ green $\rightarrow$ red $\rightarrow$ yellow — red has a longer wavelength than yellow, so putting red before yellow breaks the increasing order.
• C green $\rightarrow$ red $\rightarrow$ yellow $\rightarrow$ blue — this jumps around and ends on blue, the shortest wavelength, so it is not increasing.
• D red $\rightarrow$ yellow $\rightarrow$ green $\rightarrow$ blue — this is the order of decreasing wavelength (red longest, blue shortest), the exact reverse of what is asked.

So the correct answer is A.

Key Takeaways

• The visible spectrum is a small part of the electromagnetic spectrum.
• Wavelength increases from violet to red: violet, blue, green, yellow, orange, red.
• Frequency is the opposite: red has the lowest frequency and violet the highest, since $v = f\lambda$ and all visible light travels at the same speed $c$.
• A common memory aid is the order of colours in a rainbow: ROYGBIV (red, orange, yellow, green, blue, indigo, violet) — but that is decreasing wavelength, so read the question carefully.

Common Mistakes

• Confusing increasing and decreasing wavelength: option D is the famous rainbow order but is decreasing, so it is wrong here.
• Placing red before yellow (option B): red has a longer wavelength than yellow, not a shorter one.
• Forgetting that all electromagnetic waves, including all colours of light, travel at the same speed in a vacuum.

Things to Be Careful About

• Read whether the question wants increasing or decreasing wavelength — the same list can be correct or incorrect depending on the direction.
• Remember that wavelength and frequency are inversely related for light: longer wavelength means lower frequency.
• The visible spectrum spans roughly 400 nm (violet) to 700 nm (red); these values are useful context but not needed to answer this question.

Walkthrough

A thin converging lens acts as a magnifying glass when the object is placed between the lens and its principal focus. The rays of light emerging from the lens spread out, so they never actually meet on the far side of the lens. The image is formed where the backwards extensions of the rays meet on the same side as the object – this is a virtual image. It is upright and larger than the object.

Linear magnification tells you how many times bigger the image is than the object. Its definition is:

So the image here is virtual, and the expression for linear magnification is image length divided by object length. That matches row C.

Key Takeaways

• A converging lens used as a magnifying glass has the object inside the focal length.
• The image produced is virtual, upright and magnified.
• Linear magnification is a ratio: image length divided by object length. It has no unit.
• For the same lens, moving the object outside the focal length gives a real, inverted image instead.

Common Mistakes

• Choosing real: this is wrong because the image from a magnifying glass cannot be projected onto a screen; the rays only appear to come from the image.
• Inverting the ratio: rows B and D give object length over image length. Magnification is image over object, not object over image.
• Forgetting that the question asks for both the nature and the expression – both parts must match the same row.

Things to Be Careful About

• The phrase “magnifying glass” is a cue: object distance $u$ is less than the focal length $f$.
• If the object is placed between $F$ and the lens, the image is virtual; if placed beyond $F$, the image becomes real and inverted.
• Linear magnification can also be written as $\frac{v}{u}$ using image distance and object distance, but with lengths of object and image the expression is image length over object length.

Walkthrough

Ultrasound is sound with a frequency higher than the human ear can hear, usually above about 20 000 Hz. It is useful because it can be directed, reflected and detected, and it can carry energy into materials.

Each option is a known application of ultrasound except one:

• Cleaning jewellery is a real use. Ultrasonic cleaners send high-frequency sound waves through a liquid, producing tiny bubbles that gently remove dirt from delicate objects.
• Scanning an unborn baby is a real use. Ultrasound waves are sent into the body and the reflected waves are used to build a picture of the baby.
• Sonar is a real use. Sound waves are sent through water and the time for the echo to return tells us the distance to an object, such as the sea floor or a submarine.
• Sterilising water is not a standard use of ultrasound. Sterilising water is usually done by boiling, chemicals, or ultraviolet radiation, not by ultrasound.

So the application that is not a use for ultrasound is D.

Key Takeaways

• Ultrasound is sound, not light or other radiation.
• Ultrasound can be used for imaging, cleaning, and echo-based detection.
• Common examples include medical scanning, sonar, and ultrasonic cleaning.
• To answer a “which is not” question, check each option against the known uses and reject the odd one.

Common Mistakes

• Choosing cleaning jewellery because you do not know that ultrasonic cleaners exist. It is a genuine use of ultrasound.
• Confusing ultrasound with ultraviolet. Ultraviolet radiation is used to sterilise some things, but ultrasound is not.
• Thinking that because ultrasound can clean, it can also sterilise. Cleaning removes dirt; sterilising removes microorganisms, and that is not a standard ultrasound application.

Things to Be Careful About

• Ultrasound is a sound wave, so it needs a medium, such as air, water, or body tissue, to travel through.
• The word “ultrasound” refers to high-frequency sound, not to electromagnetic waves.
• In a “which is not” question, read every option before choosing. The obvious one is not always the answer.

Walkthrough

The potential difference (p.d.) across a component is a measure of the energy transferred to or from each coulomb of charge as it passes through the component. So:

This is exactly option C. Work done and energy transferred are the same thing here, so p.d. is also energy transferred per unit charge.

Now check the other options. Option A, voltage × current, is power: $P = VI$. Option B, energy ÷ time, is also power: $P = E/t$. Option D, current ÷ resistance, is not correct; the correct relation is Ohm's law, $V = IR$, which rearranges to $I = V/R$. So only C is correct.

Key Takeaways

• Potential difference is work done per unit charge: $V = W/Q$.
• Power is energy per unit time: $P = E/t$.
• Electrical power is also $P = VI$.
• Ohm's law relates current, p.d. and resistance: $V = IR$.
• The unit of p.d. is the volt, which is one joule per coulomb.

Common Mistakes

• Choosing A or B because they look like electrical equations; both are power, not p.d.
• Choosing D by misremembering Ohm's law; the correct form is $V = IR$, not $V = I/R$.
• Writing p.d. as energy per second instead of energy per coulomb.

Things to Be Careful About

• p.d. is energy per unit charge, not energy per unit time.
• In calculations, work done must be in joules and charge in coulombs to get p.d. in volts.
• The terms voltage and potential difference are used interchangeably in 5054.

Walkthrough

To solve this, we determine the total resistance of each circuit in terms of the resistance $R$ of a single resistor. We look at how the resistors are connected to the cell (battery).

1. Circuit 1: The two resistors are arranged one above the other. The left ends are connected together to the negative terminal, and the right ends are connected together to the positive terminal. This is a parallel connection. For two identical resistors in parallel, the total resistance is $\frac{R}{2}$.
2. Circuit 2: The wiring is drawn in a zig-zag pattern, but tracing the nodes shows that the left ends of both resistors connect to one terminal of the cell, and the right ends connect to the other. This is still a parallel connection. Total resistance = $\frac{R}{2}$.
3. Circuit 3: The resistors are drawn vertically. The top ends are connected to the positive terminal and the bottom ends to the negative terminal. This is a parallel connection. Total resistance = $\frac{R}{2}$.
4. Circuit 4: The current leaves the cell, goes through the left resistor, along the bottom wire, through the right resistor, and back to the cell. There is only one path for the current. This is a series connection. For two identical resistors in series, the total resistance is $R + R = 2R$.
5. Circuit 5: One resistor is vertical and one is horizontal. Tracing the connections: the top of the vertical resistor and the right end of the horizontal resistor connect to the positive terminal. The bottom of the vertical resistor and the left end of the horizontal resistor connect to the negative terminal. Each resistor has its own branch across the cell. This is a parallel connection. Total resistance = $\frac{R}{2}$.

Comparing the results: Circuits 1, 2, 3 and 5 have resistance $\frac{R}{2}$, while Circuit 4 has resistance $2R$. The circuits with equal resistance are 1, 2, 3 and 5.

Key Takeaways

• Series vs Parallel: In a series circuit, there is only one path for current; resistances add up ($R_{total} = R_1 + R_2$). In a parallel circuit, there are multiple paths; the total resistance is less than any individual resistance ($\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2}$).
• Circuit Diagrams: Resistors can be drawn in any orientation (horizontal, vertical, zig-zag) or arranged in non-rectangular shapes. The electrical connection (which ends are joined together) determines if they are in series or parallel, not the visual shape.
• Identical Resistors: For $n$ identical resistors of resistance $R$ in parallel, the total resistance is $\frac{R}{n}$. For $n$ in series, it is $nR$.

Common Mistakes

• Misinterpreting wiring: Students may look at Circuit 2 or 5 and assume they are series or a mix because the wires are drawn in a 'U' shape or zig-zag. Always trace the nodes (connection points) to determine the actual circuit topology.
• Confusing series and parallel resistance values: Remember that adding resistors in parallel decreases total resistance, while adding them in series increases it. Circuit 4 (series) will always have higher resistance than the parallel circuits (1, 2, 3, 5).
• Ignoring orientation: Assuming that because a resistor is drawn vertically it must be in series with a horizontal one. Check the nodes.

Things to Be Careful About

• Node identification: In diagrams like 5, look for the black dots (nodes). A dot indicates a junction where wires are connected. In Circuit 5, the top-right dot connects the positive terminal to the top of the right resistor and the wire leading to the right resistor of the bottom branch. The bottom-left dot connects the negative terminal to the left of the bottom resistor and the bottom of the right resistor. This confirms they are in parallel.
• Visual tricks: Examination boards often draw parallel circuits with 'staircase' or 'zig-zag' wires to test if students truly understand the concept or are just looking for a standard rectangular box shape.

Walkthrough

The question asks which appliance uses the heating effect of electricity. The heating effect happens when an electric current flows through a material that has resistance, such as the thin wire in a heating element. The electrical energy is transferred into thermal energy, making the element hot.

A hairdryer contains such a heating element, so it is clearly designed to use the heating effect. A fan also contains an electric motor, which uses the magnetic effect of a current to turn the blades, not the heating effect. A lawnmower also uses an electric motor to turn its blade. A cell phone uses electricity for its screen, speaker and circuits, not mainly for heating.

Key Takeaways

• The heating effect of a current is used in appliances such as hairdryers, kettles, toasters and electric heaters.
• The heating effect happens because the resistance of a wire transfers electrical energy to thermal energy when a current flows through it.
• Different appliances use different effects of electricity: heating, magnetic and chemical effects.

Common Mistakes

• Choosing a fan because it might feel warm after long use. The question asks about the appliance that uses the heating effect as its main purpose, not one that becomes warm as a side effect.
• Choosing a cell phone because it can become warm when used for a long time. Its main purpose is not heating.

Things to Be Careful About

• Read the question carefully: it asks which appliance uses the heating effect of electricity, not which appliance gets warm.
• The hairdryer is the only option that contains a heating element designed to make heat as its main job.

Walkthrough

This question asks for electrical energy in kilowatt-hours. A kilowatt-hour is the energy used by a power of 1 kilowatt running for 1 hour, so the easiest route is:

where $E$ is energy in kilowatt-hours, $P$ is power in kilowatts and $t$ is time in hours.

The heater uses $1000\ \text{W}$, which is $1.0\ \text{kW}$. The time is 30 minutes, which is $0.50\ \text{h}$. Multiplying gives:

So the correct option is B.

The 230 V supply is a distractor. The power is already given, so there is no need to use the voltage or to calculate the current.

Key Takeaways

• Electrical energy in kilowatt-hours is found from power in kilowatts multiplied by time in hours.
• Always convert watts to kilowatts and minutes to hours before using this method.
• The voltage is not needed when the power is already stated.

Common Mistakes

• Using 30 minutes as 0.3 h instead of 0.5 h, which gives 0.3 kWh and option A.
• Forgetting to convert 1000 W into 1 kW, then getting a very large number of kilowatt-hours.
• Using the 230 V supply to calculate current first, which is unnecessary and wastes time.
• Confusing kilowatt-hours (energy) with kilowatts (power).

Things to Be Careful About

• Always state the unit with a numerical answer.
• 30 minutes is half an hour, so the time in hours is 0.50, not 0.30.
• The answer in kilowatt-hours is small because the heater is only on for half an hour.

Walkthrough

An electric kettle has a metal (stainless-steel) outer casing. If the live wire becomes loose and touches the casing, the casing becomes live. If someone touches it, current could flow through them to earth, causing a severe electric shock. To prevent this, the casing is connected to the earth wire, which provides a low-resistance path to earth. The large current that flows when the casing becomes live blows the fuse, disconnecting the appliance. The fuse must be in the live wire so that when it blows, the appliance is disconnected from the live supply. A fuse in the neutral wire would not disconnect the live supply, leaving the appliance still dangerous.

Key Takeaways

• Metal-cased appliances must be earthed.
• The fuse must be in the live wire.
• Earthing works with the fuse to protect against electric shock.

Common Mistakes

• Thinking a fuse in the neutral wire is acceptable — it is not, because the live wire remains connected.
• Forgetting that a metal casing requires earthing even if a fuse is present.

Things to Be Careful About

• The earth wire is a safety wire, not part of the normal current path.
• The fuse rating should be slightly above the normal operating current.
• "Double insulation" is an alternative to earthing, but this kettle has a metal casing, so it must be earthed.

Walkthrough

The earth wire is a safety feature in a mains plug. Its job is to provide a low-resistance path for electric current to flow directly to the ground if the metal casing of an appliance becomes live (for example, because an internal wire has come loose and touched the case). When this happens, a very large current flows through the earth wire, which causes the fuse to blow or the circuit breaker to trip, cutting off the supply. This stops the casing from staying live, so a person touching it does not receive an electric shock.

Option C is therefore correct: the earth wire prevents a person getting a shock. Option A is wrong because the circuit is completed by the live and neutral wires, not the earth wire. Option B is wrong because the earth wire does not conduct thermal energy to cool the appliance. Option D is wrong because the earth wire does not protect the appliance from a large current; it protects people from a live casing.

Key Takeaways

The earth wire is a safety device that protects people from electric shock. It provides a low-resistance path for current to the ground if the metal casing becomes live, causing the fuse to blow or the circuit breaker to trip. The earth wire does not complete the circuit, does not cool the appliance, and does not protect the appliance itself.

Common Mistakes

• Choosing A: thinking the earth wire completes the circuit. It does not; the live and neutral wires do that.
• Choosing D: confusing earthing with fuse protection. The fuse protects the appliance from a large current, but the earth wire protects people from a live casing.
• Choosing B: thinking the earth wire conducts heat, which is not its purpose.

Things to Be Careful About

The earth wire is only present in appliances with metal casings. Double-insulated appliances have a plastic casing and do not need an earth wire. The key idea is that the earth wire creates a path for current to flow to the ground, which makes the fuse blow and prevents the casing from staying live.

Walkthrough

The question asks for the direction of the force on a current-carrying wire placed in a magnetic field. This is determined using Fleming's left-hand rule.

1. Identify the direction of the magnetic field: the field lines point from left to right, so the first finger points to the right.
2. Identify the direction of the current: the arrow on the wire points upwards, so the second finger points upwards.
3. Orient the left hand so that the first finger points right and the second finger points up. The thumb will naturally point into the page (away from the viewer).
4. The thumb represents the direction of the force (or motion) on the conductor. Therefore, the force is into the page.

Key Takeaways

• Fleming's left-hand rule is used to find the direction of the force on a current-carrying conductor in a magnetic field.
• First finger = magnetic Field, Second finger = current (conventional), Thumb = Thrust (Force/motion).
• The three directions (force, field, current) are mutually perpendicular.

Common Mistakes

• Using the right hand instead of the left hand (which gives the direction of induced current in electromagnetic induction, not force on a conductor).
• Confusing conventional current (positive to negative) with electron flow (negative to positive); Fleming's left-hand rule uses conventional current.
• Misaligning the fingers: the first and second fingers must be at right angles to each other.

Things to Be Careful About

• Ensure you are using the LEFT hand for the motor effect (force on a conductor) and the RIGHT hand for the generator effect (induced current). A common mnemonic is "Left = Motor, Right = Generator".
• The current direction is the conventional current (positive to negative), as shown by the upward arrow in the diagram.

Walkthrough

The diagram shows a simple d.c. motor. The key components are the magnets (creating the magnetic field), the coil (carrying the current and experiencing a force), the brushes (making contact with the rotating part), and the split-ring commutator (reversing the current in the coil every half-turn to keep the coil rotating in the same direction). Part Q is the split circular ring that is cut into two halves, which is the split-ring commutator. A slip ring would be a complete, unbroken ring, used in an a.c. generator. Therefore, Q is a split-ring commutator.

Key Takeaways

• A simple d.c. motor uses a split-ring commutator to reverse the current in the coil every half-turn, ensuring continuous rotation in one direction.
• A slip ring (complete ring) is used in an a.c. generator to maintain continuous contact without reversing the current.

Common Mistakes

• Confusing a split-ring commutator with a slip ring. The split ring has a gap, making it two separate halves.

Things to Be Careful About

• Look closely at the ring: if it has a gap, it is a split-ring commutator (d.c. motor). If it is a complete continuous ring, it is a slip ring (a.c. generator).