ESAT Paper 2 Physics Sample Questions

Correct Answer: C. 2

Step-by-Step Breakdown:

1. Add up the conductor inside the field

Each of the 5 strands lies in the field over $0.25m$, and the bends that join them sit outside the field, so they contribute nothing. The current runs the same way in every strand, so the forces on the strands act in the same direction and add, and the length to use is the total length inside the field:
$$L=5\times 0.25=1.25m.$$

2. Make the current the subject

$$F=BIL\Rightarrow I=\frac{F}{BL}.$$

3. Substitute and check

$$I=\frac{3}{1.2\times 1.25}=\frac{3}{1.5}=2A.$$

Working forwards as a check: at $2A$ one strand feels $1.2\times 2\times 0.25=0.6N$, and 5 identical strands give $3N$, the force stated.

The key is $2A$.

Why the Other Options Are Wrong (❌):

• A. 10 · Only one strand counted
  Treats a single crossing of the field as the whole conductor, $L=0.25m$: $3÷(1.2\times 0.25)=10A$. That is the current one strand would need on its own, and the wire makes 5 crossings of the field.
• B. 0.4 · Strand count applied twice
  Shares the force between the 5 strands and then uses the total length as well, so the folding is counted twice: $3÷5=0.6N$, then $0.6÷(1.2\times 1.25)=0.4A$. Either share the force and use one strand's $0.25m$, or keep the whole force and use $1.25m$: both give $2A$.
• E. 2.5 · Length omitted
  Divides the force by the flux density alone: $3÷1.2=2.5A$. 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 $1.25m$ to be taken out of it.
• G. 20 · Decade lost in the division
  The right route with the decimal point a place out: $3÷1.5$ is $2A$, 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 $0.6N$ and the 5 strands together ten times $3N$, not the $3N$ the question states.
• H. 50 · Strand count applied to the force instead of the length
  The folding is worth a factor of 5, and this route puts that factor on the force rather than on the length: $3\times 5$ divided by $1.2\times 0.25$. The same figure comes of working out the current a single strand would need, $10A$, and then multiplying by 5 again because there are 5 strands. Both moves push the wrong way. The 5 crossings make the conductor in the field $1.25m$ long, so the current needed for $3N$ falls by a factor of 5 below $10A$, to $2A$.

Common Mistake (⚠️):
Using one crossing of the field as though a single conductor were in it. The wire makes 5 crossings of the field with the current the same way each time, so the conductor in the field is $1.25m$ rather than $0.25m$, and the current needed for the stated force is smaller than a single strand would need by a factor of 5.

Takeaway (📌):
$F=BIL$ counts metres of conductor inside the field, not wires. Fold a wire so that 5 strands cross the same gap carrying the current the same way and the length to use is $5\times 0.25=1.25m$, so the current needed for a fixed force is a factor of 5 lower than one strand would need.

Correct Answer: E. 3

Step-by-Step Breakdown:

1. Add the distances, then add the times

The journey covers $600m$ before the wait and $300m$ after it, so the total distance is

$$600+300=900m.$$

The clock runs for $120s$, then $60s$, then $120s$. The middle stage adds no distance, but it does add time, and an average taken over the whole journey has to include it:

$$120+60+120=300s.$$

2. Divide the total distance by the total time

$$\text{average speed}={\displaystyle \frac{900m}{300s}}$$$$=3\mathrm{m}{\mathrm{s}}^{-1}$$

Check: dividing the same $900m$ by the $240s$ actually spent moving would give $3.75\mathrm{m}{\mathrm{s}}^{-1}$, 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 $3$.

Why the Other Options Are Wrong (❌):

• A. 1 · Counts only the distance travelled after the wait
  Takes only the distance covered after the wait, $300m$, and divides it by the full $300s$: $300÷300=1$. The $600m$ covered before the wait is part of the journey too, so the numerator is short by exactly that much.
• B. 2.5 · Quotes the speed of the final stage
  The last stage covers $300m$ in $120s$, so $300÷120=2.5$. That is the speed for that stage alone, and an average over the whole journey has to take in the first stage and the $60s$ wait as well.
• C. 3.75 · Leaves the stopped time out of the total time
  Uses the moving time only, $120+120=240s$, giving $900÷240=3.75$. That is the average speed while moving, a real quantity but not the one asked for: the $60s$ spent at rest still belongs in the total time.
• D. 5 · Quotes the speed of the first stage
  The first stage covers $600m$ in $120s$, so $600÷120=5$. That is the speed of one stage rather than an average: the journey runs on for another $60s$ of waiting and $120s$ of motion after it.
• F. 7.5 · Stage speeds added as though speeds combined like distances
  Works out each stage on its own, $600÷120=5$ and $300÷120=2.5$, and then adds the two: $5+2.5$. Distances add and times add, and that is why the totals $900m$ and $300s$ 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 $900÷300=3$.

Common Mistake (⚠️):
Stopping the clock during the wait. Dividing the total $900m$ by only the $240s$ actually spent moving gives $3.75\mathrm{m}{\mathrm{s}}^{-1}$, which is the average speed while moving. The question asks for the average over the whole journey, and the $60s$ spent at a field gate is part of that journey.

Takeaway (📌):
An average speed is not the average of the speeds. Add the distances, add the times, divide once, and leave the wait sitting inside the total time where it belongs.

Correct Answer: D. $8{\text{m s}}^{-2}$ downwards

Step-by-Step Breakdown:

1. Work in newtons per kilogram

Gravity supplies $10\text{N}$ to every kilogram of the bag. The drag acts upwards against the weight, and per kilogram it comes to

$$\frac{D}{m}=\frac{18}{9}=2{\text{N kg}}^{-1}$$

2. Take the difference and read it as an acceleration

What is left is unbalanced, and one newton per kilogram accelerates one kilogram at one metre per second squared:

$$a=10-2=8{\text{m s}}^{-2}$$

The long way agrees: $W=9\times 10=90\text{N}$, then $F=90-18=72\text{N}$ downwards, and $72÷9=8$.

The key is $8{\text{m s}}^{-2}$ downwards.

Why the Other Options Are Wrong (❌):

• A. $2{\text{m s}}^{-2}$ downwards · Weight left out of the resultant
  Takes the drag reading to be the whole resultant force and divides that by the mass: $18÷9=2$. The weight is the larger of the two forces on the bag, and $18\text{N}$ is only the part of it the air is carrying, so it cannot stand in for the resultant.
• B. $12{\text{m s}}^{-2}$ downwards · Opposing forces added
  Adds the two vertical forces instead of opposing them: $(90+18)÷9=108÷9=12$. 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 $90\text{N}$ weight.
• C. $10{\text{m s}}^{-2}$ downwards · Drag ignored, free fall assumed
  Ignores the drag and lets the bag fall freely, so $a=g=10$. Free fall would need the air to push with no force at all, and the question puts $18\text{N}$ of drag on the bag.
• E. $7.2{\text{m s}}^{-2}$ downwards · Resultant divided by the field strength
  Divides the resultant by the field strength rather than by the mass, $72÷10$, which gives $7.2{\text{m s}}^{-2}$ downwards. Dividing a force by $10{\text{N kg}}^{-1}$ returns a mass in kilograms; Newton's second law divides by the $9\text{kg}$ mass, and only that turns a resultant into an acceleration.
• F. $0.8{\text{m s}}^{-2}$ downwards · Resultant divided by the weight in newtons
  Divides the resultant by the weight rather than by the mass: $\frac{72}{90}$, which is $\frac{8}{10}$ and so carries the right digits with the decimal point one place out. Newton's second law divides by the $9\text{kg}$ mass, and $90\text{N}$ is that mass already multiplied by $10$, so dividing by it removes the field strength a second time and leaves an acceleration $10$ times too small.
• G. $9.8{\text{m s}}^{-2}$ downwards · Drag shared over the weight instead of the mass
  Uses the per-kilogram route but shares the drag over the weight: $\frac{18}{90}$ instead of $\frac{18}{9}$, then takes it off $10$. That leaves the fall looking almost free, just under $10{\text{m s}}^{-2}$, because dividing by $90\text{N}$ shrinks the drag by a further factor of $10$. The drag is shared over the $9\text{kg}$ that is being accelerated, so it is $2\text{N}$ per kilogram, and $10-2=8$.

Common Mistake (⚠️):
Adding the drag to the weight because both act on the same body, giving $90+18=108\text{N}$ and an acceleration of $12{\text{m s}}^{-2}$. The air pushes upwards while gravity pulls downwards, so the two must be subtracted, and the resultant can never be larger than the $90\text{N}$ weight.

Takeaway (📌):
Divide the drag by the mass before anything else. The field hands every kilogram $10\text{N}$, the air takes $2$ of that back, and the difference $10-2=8$ is the acceleration in ${\text{m s}}^{-2}$, so the weight never has to be worked out at all.

Correct Answer: G. 1

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

$$f=3\times 250=750\text{Hz}$$

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

The wave travels at $1500{\text{m s}}^{-1}$, a speed set by the medium and not by the source, so

$$\lambda =\frac{v}{f}=\frac{1500}{750}=2\text{m}$$

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 a compression to the nearest rarefaction, and that is half a wavelength:

$$d=1\text{m}$$

The key is $1\text{m}$.

Why the Other Options Are Wrong (❌):

• A. 2 · The other of the two spacings quoted
  This is the distance from one compression to the next compression, which is one whole wavelength: $2m$. The question names the distance from a compression to the nearest rarefaction, and that is half a wavelength, $1m$. One of the two is twice the other, so they cannot both answer the same question.
• B. 3 · The reference frequency used in place of the source's
  Puts the reference $250\mathrm{H}\mathrm{z}$ into the wave equation rather than the frequency of the source itself. That gives $1500÷250=6m$ as the wavelength, which makes the compression to rarefaction spacing $3m$ instead of $1m$. The source is running at three times the reference frequency, $750\mathrm{H}\mathrm{z}$, and it is that figure the wavelength follows from.
• C. 0.5 · A quarter of a cycle taken
  Divides the wavelength into four, $2÷4=0.5m$. 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.
• D. 9 · Wavelength scaled in step with the frequency rather than against it
  Starts from the reference wavelength, $1500÷250=6m$, then multiplies it by the same factor that was applied to the frequency before taking the compression to rarefaction spacing, which gives $9m$. At a fixed wave speed the wavelength follows the reciprocal of that factor, not the factor itself: the wave equation applied directly gives $\lambda =1500÷750=2m$.
• E. 6 · Reference frequency used and the other spacing taken from it
  Two slips at once. The reference $250\mathrm{H}\mathrm{z}$ goes into the wave equation in place of the source's own $750\mathrm{H}\mathrm{z}$, giving $1500÷250=6m$, and the distance then read off it is one whole wavelength, the spacing from one compression to the next compression. The source runs at three times the reference frequency, so the wavelength to use is $1500÷750=2m$, and the question asks from a compression to the nearest rarefaction, which is half a wavelength: $1m$.
• F. 27 · Frequency factor multiplied into the wavelength twice over
  Starts from the reference wavelength $1500÷250=6m$ and multiplies by the frequency factor, then multiplies by it again before taking half a 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 $\lambda =1500÷750=2m$, and half a wavelength of that is $1m$.

Common Mistake (⚠️):
Answering with the distance from one compression to the next compression instead. That spacing is one whole wavelength, $2m$, while the question names the distance from a compression to the nearest rarefaction, which is half a wavelength, $1m$. 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.