ESAT Paper 3 Physics Sample Questions

Correct Answer: D. 17.5

Step-by-Step Breakdown:

1. Use the symbols to decide which branches carry current

The battery's positive terminal is joined to the left-hand rail, so conventional current leaves that terminal, runs along the left-hand rail and crosses from left to right through any branch that conducts before returning down the right-hand rail. A diode passes current only in the direction its triangle points.

The triangle in the $1\ \Omega$ branch points the way the current is trying to go, so that branch conducts as though the diode were a plain wire. The triangle in the $7\ \Omega$ branch points against the flow, so that branch is an open circuit and carries nothing whatever. The $4\ \Omega$ branch holds no diode, so it always conducts.

2. Add the currents in the branches that survive

Each live branch is connected straight across the supply, so the full $14\ \text{V}$ appears across it:
$$I_1 = \frac{14}{1} = 14.0\ \text{A}, \qquad I_2 = \frac{14}{4} = 3.5\ \text{A}, \qquad I_3 = 0.$$

The ammeter sits in the main lead, so it records everything leaving the battery:
$$I = 14.0 + 3.5 = 17.5\ \text{A}.$$

Sanity check by the other route: the two live branches combine to $\frac{1 \times 4}{1 + 4} = 0.8\ \Omega$, and $14 \div 0.8 = 17.5$, which agrees.

The key is $17.5$.

Why the Other Options Are Wrong (❌):

• A. 19.5 · Ignored the diodes
  Lets all three branches conduct, as though the diodes were not there: $14.0 + 3.5 + 2.0 = 19.5$. The branch whose diode opposes the current is an open circuit and adds nothing to the reading.
• B. 5.5 · Diode directions read the wrong way round
  Takes the $7\ \Omega$ branch to be the conducting one and the $1\ \Omega$ branch to be blocked: $3.5 + 2.0 = 5.5$. A diode passes current in the direction its triangle points, and it is the triangle in the $1\ \Omega$ branch that points the way the current is going.
• C. 3.5 · Diode taken to block in both directions
  Deletes both diode branches, leaving only the plain $4\ \Omega$ branch: $14 \div 4 = 3.5$. A diode blocks one way only, so the $1\ \Omega$ branch is still carrying current.
• E. 0.8 · Stopped at the combined resistance
  Combines the two conducting branches correctly, $\frac{1 \times 4}{1 + 4} = 0.8$, and writes that number down. It is a resistance in ohms, not a current, and one division is still owed: $14 \div 0.8 = 17.5$.
• F. 11.2 · Combined resistance multiplied by the supply instead of divided into it
  The two conducting branches are combined correctly, $\frac{1 \times 4}{1 + 4} = 0.8\ \Omega$, and the supply is then multiplied by that resistance instead of divided by it: $14 \times 0.8$. Ohm's law gives a current as a voltage divided by a resistance, so the last step runs the other way, $14 \div 0.8 = 17.5\ \text{A}$, which is the same number as adding the branch currents $14.0$ and $3.5$.
• G. 245.0 · Power formula used in place of Ohm's law
  Each live branch is worked out with $\dfrac{V^{2}}{R}$, which is the power it dissipates, rather than with $\dfrac{V}{R}$: $\dfrac{14^{2}}{1} + \dfrac{14^{2}}{4}$. That total is in watts and the ammeter reads amperes. One factor of $14$ too many has been used: the branch currents are $14 \div 1 = 14.0$ and $14 \div 4 = 3.5$, so the meter shows $17.5\ \text{A}$.
• H. 1.25 · Supply voltage never applied
  The reciprocals of the two conducting resistances are added, $\dfrac{1}{1} + \dfrac{1}{4}$, and that sum is written down as the reading. It is a combined conductance in siemens and it makes no use of the supply at all. A branch current is the supply voltage divided by the branch resistance, $14 \div 1 = 14.0$ and $14 \div 4 = 3.5$, and the ammeter in the main lead reads their sum, $17.5\ \text{A}$.

Common Mistake (⚠️):
Treating a diode as though it merely reduced the current in its branch rather than removing that branch from the circuit. An ideal reverse biased diode is a break in the wire, so the $7\ \Omega$ branch contributes exactly zero and not a reduced share, and the reading is $14.0 + 3.5$ and nothing else.

Takeaway (📌):
Direction first, arithmetic second. Mark the way conventional current leaves the positive terminal, cross out any branch whose diode points against it, then divide the supply voltage by each surviving resistance and add. A diode question is a reading task with a short sum on the end of it.

Correct Answer: D. 1.5

Step-by-Step Breakdown:

1. Find the length of conductor in the field

The field acts on $0.2\ \mathrm{m}$ of each strand and on nothing else, because the rest of each turn lies outside it. Every strand carries the current in the same direction, so their forces point the same way and add, and the arrangement behaves as one conductor of length
$$L = 6 \times 0.2 = 1.2\ \mathrm{m}.$$

2. Rearrange $F = BIL$

With the current perpendicular to the field, $F = BIL$, so
$$I = \frac{F}{BL}.$$

3. Put the numbers in

The denominator first: $1.5 \times 1.2 = 1.8$. Then
$$I = \frac{2.7}{1.8} = 1.5\ \mathrm{A}.$$

Checking backwards, one strand at $1.5\ \mathrm{A}$ feels $1.5 \times 1.5 \times 0.2 = 0.45\ \mathrm{N}$, and 6 of them give the stated $2.7\ \mathrm{N}$.

The key is $1.5\ \mathrm{A}$.

Why the Other Options Are Wrong (❌):

• A. 9 · Only one strand counted
  Treats a single crossing of the field as the whole conductor, $L = 0.2\ \mathrm{m}$: $2.7 \div (1.5 \times 0.2) = 9\ \mathrm{A}$. That is the current one strand would need on its own, and the wire makes 6 crossings of the field.
• B. 0.25 · Strand count applied twice
  Shares the force between the 6 strands and then uses the total length as well, so the factor of 6 is counted twice: $2.7 \div 6 = 0.45\ \mathrm{N}$, then $0.45 \div (1.5 \times 1.2) = 0.25\ \mathrm{A}$. Either share the force and use one strand's $0.2\ \mathrm{m}$, or keep the whole force and use $1.2\ \mathrm{m}$: both give $1.5\ \mathrm{A}$.
• C. 2.25 · Flux density omitted
  Divides the force by the length alone and leaves the flux density out of the rearrangement: $2.7 \div 1.2 = 2.25\ \mathrm{A}$. There are three factors on the right of $F = BIL$, so making $I$ the subject divides by $1.5\ \mathrm{T}$ as well as by $1.2\ \mathrm{m}$.
• E. 1.8 · Length omitted
  Divides the force by the flux density alone: $2.7 \div 1.5 = 1.8\ \mathrm{A}$. The length in the field has to divide as well. The units settle it: newtons per tesla is an ampere multiplied by a metre, not an ampere, so this figure still has $1.2\ \mathrm{m}$ to be taken out of it.
• F. 0.45 · Force on one strand reported instead of the current
  Shares the total force between the 6 strands and stops there: $2.7 \div 6 = 0.45\ \mathrm{N}$. That figure is a force, and it is the one the solution works out as its own check. Turning it into a current still needs the two divisions that make $I$ the subject, by $1.5\ \mathrm{T}$ and by the $0.2\ \mathrm{m}$ that one strand lies in the field, which gives $1.5\ \mathrm{A}$.
• G. 15 · Decade lost in the division
  The right route with the decimal point a place out: $2.7 \div 1.8$ is $1.5\ \mathrm{A}$, and this figure is ten times it. Working the physics forwards catches a slipped decade, because a current ten times too large gives ten times the force: one strand would feel ten times $0.45\ \mathrm{N}$ and the 6 strands together ten times $2.7\ \mathrm{N}$, not the $2.7\ \mathrm{N}$ the question states.
• H. 54 · Strand count applied to the force instead of the length
  The 6 strands are worth a factor of 6, and this route puts that factor on the force rather than on the length: $2.7 \times 6$ divided by $1.5 \times 0.2$. The same figure comes of working out the current a single strand would need, $9\ \mathrm{A}$, and then multiplying by 6 again because there are 6 strands. Both moves push the wrong way. The 6 crossings make the conductor in the field $1.2\ \mathrm{m}$ long, so the current needed for $2.7\ \mathrm{N}$ falls by a factor of 6 below $9\ \mathrm{A}$, to $1.5\ \mathrm{A}$.

Common Mistake (⚠️):
Using one crossing of the field as though a single conductor were in it. The wire makes 6 crossings of the field with the current the same way each time, so the conductor in the field is $1.2\ \mathrm{m}$ rather than $0.2\ \mathrm{m}$, and the current needed for the stated force is smaller than a single strand would need by a factor of 6.

Takeaway (📌):
$F = BIL$ counts metres of conductor inside the field, not wires. Wind a wire into a coil so that 6 strands cross the same gap carrying the current the same way and the length to use is $6 \times 0.2 = 1.2\ \mathrm{m}$, so the current needed for a fixed force is a factor of 6 lower than one strand would need.

Correct Answer: F. 1.5

Step-by-Step Breakdown:

1. Add the distances, then add the times

The journey covers $150\,\mathrm{m}$ before the wait and $300\,\mathrm{m}$ after it, so the total distance is

The clock runs for $60\,\mathrm{s}$, then $180\,\mathrm{s}$, then $60\,\mathrm{s}$. The middle stage adds no distance, but it does add time, and an average taken over the whole journey has to include it:

2. Divide the total distance by the total time

Check: dividing the same $450\,\mathrm{m}$ by the $120\,\mathrm{s}$ actually spent moving would give $3.75\,\mathrm{m\,s^{-1}}$, which is larger because the same distance is being spread over less time. That figure is the average speed while moving, not the average asked for.

The key is $1.5$.

Why the Other Options Are Wrong (❌):

• A. 1 · Counts only the distance travelled after the wait
  Takes only the distance covered after the wait, $300\,\mathrm{m}$, and divides it by the full $300\,\mathrm{s}$: $300 \div 300 = 1$. The $150\,\mathrm{m}$ covered before the wait is part of the journey too, so the numerator is short by exactly that much.
• B. 5 · Quotes the speed of the final stage
  The last stage covers $300\,\mathrm{m}$ in $60\,\mathrm{s}$, so $300 \div 60 = 5$. That is the speed for that stage alone, and an average over the whole journey has to take in the first stage and the $180\,\mathrm{s}$ wait as well.
• C. 3.75 · Leaves the stopped time out of the total time
  Uses the moving time only, $60 + 60 = 120\,\mathrm{s}$, giving $450 \div 120 = 3.75$. That is the average speed while moving, a real quantity but not the one asked for: the $180\,\mathrm{s}$ spent at rest still belongs in the total time.
• D. 2.5 · Quotes the speed of the first stage
  The first stage covers $150\,\mathrm{m}$ in $60\,\mathrm{s}$, so $150 \div 60 = 2.5$. That is the speed of one stage rather than an average: the journey runs on for another $180\,\mathrm{s}$ of waiting and $60\,\mathrm{s}$ of motion after it.
• E. 7.5 · Stage speeds added as though speeds combined like distances
  Works out each stage on its own, $150 \div 60 = 2.5$ and $300 \div 60 = 5$, and then adds the two: $2.5 + 5$. Distances add and times add, and that is why the totals $450\,\mathrm{m}$ and $300\,\mathrm{s}$ are formed first; speeds do not add, and a figure above both stage speeds cannot be an average of them. The one division that is wanted is $450 \div 300 = 1.5$.

Common Mistake (⚠️):
Quoting one stage instead of averaging the journey. The first stage runs at $2.5\,\mathrm{m\,s^{-1}}$ and the last at $5\,\mathrm{m\,s^{-1}}$, and neither is the answer: an average speed is taken over the total distance and the total time in a single division, never read off one stage of the journey.

Takeaway (📌):
For an average speed, use the two totals and nothing else: every metre travelled over every second elapsed. A wait adds seconds and no metres, so it belongs in the denominator and nowhere else.

Correct Answer: E. $6\ \text{m s}^{-2}$ downwards

Step-by-Step Breakdown:

1. Set the two forces against each other

Only two forces act on the pipe. The weight pulls down, $W = mg = 70 \times 10 = 700\ \text{N}$, and the drag pushes up with $280\ \text{N}$. Opposite directions means subtraction, so the resultant is

directed downwards, since the weight is the larger.

2. Apply Newton's second law

The mass divides, not the weight and not $10$: dividing $420\ \text{N}$ by $10\ \text{N kg}^{-1}$ would give a number of kilograms, which is not what the question asks for.

The key is $6\ \text{m s}^{-2}$ downwards.

Why the Other Options Are Wrong (❌):

• A. $4\ \text{m s}^{-2}$ downwards · Weight left out of the resultant
  Takes the drag reading to be the whole resultant force and divides that by the mass: $280 \div 70 = 4$. The weight is the larger of the two forces on the pipe, and $280\ \text{N}$ is only the part of it the air is carrying, so it cannot stand in for the resultant.
• B. $14\ \text{m s}^{-2}$ downwards · Opposing forces added
  Adds the two vertical forces instead of opposing them: $(700 + 280) \div 70 = 980 \div 70 = 14$. Adding is right only for forces pointing the same way, and the drag points opposite to the weight, so the resultant can never be larger than the $700\ \text{N}$ weight.
• C. $10\ \text{m s}^{-2}$ downwards · Drag ignored, free fall assumed
  Ignores the drag and lets the pipe fall freely, so $a = g = 10$. Free fall would need the air to push with no force at all, and the question puts $280\ \text{N}$ of drag on the pipe.
• D. $42\ \text{m s}^{-2}$ downwards · Resultant divided by the field strength
  Divides the resultant by the field strength rather than by the mass, $420 \div 10$, which gives $42\ \text{m s}^{-2}$ downwards. Dividing a force by $10\ \text{N kg}^{-1}$ returns a mass in kilograms; Newton's second law divides by the $70\ \text{kg}$ mass, and only that turns a resultant into an acceleration.
• F. $0.6\ \text{m s}^{-2}$ downwards · Resultant divided by the weight in newtons
  Divides the resultant by the weight rather than by the mass: $\frac{420}{700}$, which is $\frac{6}{10}$ and so carries the right digits with the decimal point one place out. Newton's second law divides by the $70\ \text{kg}$ mass, and $700\ \text{N}$ is that mass already multiplied by $10$, so dividing by it also divides out the field strength and leaves an acceleration $10$ times too small.
• G. $9.6\ \text{m s}^{-2}$ downwards · Drag shared over the weight instead of the mass
  Uses the per-kilogram route but shares the drag over the weight: $\frac{280}{700}$ instead of $\frac{280}{70}$, then takes it off $10$. That leaves the fall looking almost free, just under $10\ \text{m s}^{-2}$, because dividing by $700\ \text{N}$ shrinks the drag by a further factor of $10$. The drag is shared over the $70\ \text{kg}$ that is being accelerated, so it is $4\ \text{N}$ per kilogram, and $10 - 4 = 6$.

Common Mistake (⚠️):
Adding the drag to the weight because both act on the same body, giving $700 + 280 = 980\ \text{N}$ and an acceleration of $14\ \text{m s}^{-2}$. The air pushes upwards while gravity pulls downwards, so the two must be subtracted, and the resultant can never be larger than the $700\ \text{N}$ weight.

Takeaway (📌):
Divide the drag by the mass before anything else. The field hands every kilogram $10\ \text{N}$, the air takes $4$ of that back, and the difference $10 - 4 = 6$ is the acceleration in $\text{m s}^{-2}$, so the weight never has to be worked out at all.

Correct Answer: A. 0.2

Step-by-Step Breakdown:

1. Turn the reference frequency into the frequency of this wave

The source runs at three times the reference frequency, so

2. Use $v = f\lambda$ to get one whole wavelength

The wave travels at $2.4\ \text{m s}^{-1}$, a speed set by the medium and not by the source, so

That is the distance from one compression to the next compression, because the whole pattern repeats once per wavelength.

3. Decide which fraction of a wavelength the question names

Along the direction of travel the features run compression, rarefaction, compression, evenly spaced, so a rarefaction sits exactly halfway between neighbouring compressions. The question asks for the shortest distance from one compression to the next compression, and that is one whole wavelength:

The key is $0.2\ \text{m}$.

Why the Other Options Are Wrong (❌):

• B. 0.1 · The other of the two spacings quoted
  This is the distance from a compression to the nearest rarefaction, which is half a wavelength: $0.1\,\mathrm{m}$. The question names the distance from one compression to the next compression, and that is one whole wavelength, $0.2\,\mathrm{m}$. One of the two is twice the other, so they cannot both answer the same question.
• C. 0.6 · The reference frequency used in place of the source's
  Puts the reference $4\,\mathrm{Hz}$ into the wave equation rather than the frequency of the source itself. That gives $2.4 \div 4 = 0.6\,\mathrm{m}$ as the wavelength, which makes the compression to compression spacing $0.6\,\mathrm{m}$ instead of $0.2\,\mathrm{m}$. The source is running at three times the reference frequency, $12\,\mathrm{Hz}$, and it is that figure the wavelength follows from.
• D. 0.05 · A quarter of a cycle taken
  Divides the wavelength into four, $0.2 \div 4 = 0.05\,\mathrm{m}$. A quarter of a wavelength past a compression lands midway between that compression and the rarefaction beyond it, which is neither of the two features the question names.
• E. 1.8 · Wavelength scaled in step with the frequency rather than against it
  Starts from the reference wavelength, $2.4 \div 4 = 0.6\,\mathrm{m}$, then multiplies it by the same factor that was applied to the frequency before taking the compression to compression spacing, which gives $1.8\,\mathrm{m}$. At a fixed wave speed the wavelength follows the reciprocal of that factor, not the factor itself: the wave equation applied directly gives $\lambda = 2.4 \div 12 = 0.2\,\mathrm{m}$.
• F. 0.3 · Reference frequency used and the other spacing taken from it
  Two slips at once. The reference $4\,\mathrm{Hz}$ goes into the wave equation in place of the source's own $12\,\mathrm{Hz}$, giving $2.4 \div 4 = 0.6\,\mathrm{m}$, and the distance then read off it is half a wavelength, the spacing from a compression to the nearest rarefaction. The source runs at three times the reference frequency, so the wavelength to use is $2.4 \div 12 = 0.2\,\mathrm{m}$, and the question asks from one compression to the next compression, which is one whole wavelength: $0.2\,\mathrm{m}$.
• G. 5.4 · Frequency factor multiplied into the wavelength twice over
  Starts from the reference wavelength $2.4 \div 4 = 0.6\,\mathrm{m}$ and multiplies by the frequency factor, then multiplies by it again before taking one whole wavelength. The factor belongs there once, and as a divisor: at a fixed wave speed the wavelength follows the reciprocal of a frequency change. One application, the right way round, is $\lambda = 2.4 \div 12 = 0.2\,\mathrm{m}$, and one whole wavelength of that is $0.2\,\mathrm{m}$.

Common Mistake (⚠️):
Answering with the distance from a compression to the nearest rarefaction instead. That spacing is half a wavelength, $0.1\,\mathrm{m}$, while the question names the distance from one compression to the next compression, which is one whole wavelength, $0.2\,\mathrm{m}$. One of the two is twice the other, so which pair of features the stem names settles the answer before any arithmetic does.

Takeaway (📌):
Neighbouring compressions are one wavelength apart and the rarefaction between them sits at half of that. Get $\lambda = v/f$ first, from the frequency of the source and not from whatever reference the stem quotes, then let the two feature words tell you whether the gap named is a whole wavelength or half of one.