ESAT Physics Practice Questions by Topic

Correct Answer: A. 5.0

Fastest Approach (🚀):
The reading is the sum of two branch currents, not three, so the whole question is $8 \div 2$ plus $8 \div 8$ once the blocked branch has been struck out. Any option that needs a combined resistance to reach it has cost you time you did not need to spend.

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 every branch from left to right before returning down the right-hand rail. A diode passes current only in the direction its triangle points.

The triangle in the $2\ \Omega$ 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 $16\ \Omega$ branch points against the flow, so that branch is an open circuit and carries nothing whatever. The $8\ \Omega$ 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 $8\ \text{V}$ appears across it:
$$I_1 = \frac{8}{2} = 4.0\ \text{A}, \qquad I_2 = \frac{8}{8} = 1.0\ \text{A}, \qquad I_3 = 0.$$

The ammeter sits in the main lead, so it records everything leaving the battery:
$$I = 4.0 + 1.0 = 5.0\ \text{A}.$$

Sanity check by the other route: the two live branches combine to $\frac{2 \times 8}{2 + 8} = 1.6\ \Omega$, and $8 \div 1.6 = 5.0$, which agrees.

The key is $5.0$.

Why the Other Options Are Wrong (❌):

• B. 5.5 · Ignored the diodes
  Lets all three branches conduct, as though the diodes were not there: $4.0 + 1.0 + 0.5 = 5.5$. The branch whose diode opposes the current is an open circuit and adds nothing to the reading.
• C. 1.5 · Diode directions read the wrong way round
  Takes the $16\ \Omega$ branch to be the conducting one and the $2\ \Omega$ branch to be blocked: $1.0 + 0.5 = 1.5$. A diode passes current in the direction its triangle points, and it is the triangle in the $2\ \Omega$ branch that points the way the current is going.
• D. 1.0 · Diode taken to block in both directions
  Deletes both diode branches, leaving only the plain $8\ \Omega$ branch: $8 \div 8 = 1.0$. A diode blocks one way only, so the $2\ \Omega$ branch is still carrying current.
• E. 1.6 · Stopped at the combined resistance
  Combines the two conducting branches correctly, $\frac{2 \times 8}{2 + 8} = 1.6$, and writes that number down. It is a resistance in ohms, not a current, and one division is still owed: $8 \div 1.6 = 5.0$.

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 $16\ \Omega$ branch contributes exactly zero and not a reduced share, and the reading is $4.0 + 1.0$ and nothing else.

Takeaway (📌):
Rub out every branch whose diode opposes the current before you calculate anything. What is left is a set of resistors straight across the supply, so each branch current is the supply voltage divided by that branch's resistance, and the ammeter in the main lead reads their sum.

Correct Answer: D. 2.5

Fastest Approach (🚀):
Multiply the flux density by the TOTAL length first, $1.1 \times 1.12 = 1.232$, and only then divide the force by it. One division does the whole question, and the strand count never has to be applied to the force at all, which is where the marks usually go.

Step-by-Step Breakdown:

1. Find the length of conductor in the field

The field acts on $0.28\ \mathrm{m}$ 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
$$L = 4 \times 0.28 = 1.12\ \mathrm{m}.$$

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: $1.1 \times 1.12 = 1.232$. Then
$$I = \frac{3.08}{1.232} = 2.5\ \mathrm{A}.$$

Checking backwards, one strand at $2.5\ \mathrm{A}$ feels $1.1 \times 2.5 \times 0.28 = 0.77\ \mathrm{N}$, and 4 of them give the stated $3.08\ \mathrm{N}$.

The key is $2.5\ \mathrm{A}$.

Why the Other Options Are Wrong (❌):

• A. 10 · Only one strand counted
  Treats a single crossing of the field as the whole conductor, $L = 0.28\ \mathrm{m}$: $3.08 \div (1.1 \times 0.28) = 10\ \mathrm{A}$. That is the current one strand would need on its own, and the wire makes 4 crossings of the field.
• B. 0.625 · Strand count applied twice
  Shares the force between the 4 strands and then uses the total length as well, so the strand count is applied twice: $3.08 \div 4 = 0.77\ \mathrm{N}$, then $0.77 \div (1.1 \times 1.12) = 0.625\ \mathrm{A}$. Either share the force and use one strand's $0.28\ \mathrm{m}$, or keep the whole force and use $1.12\ \mathrm{m}$: both give $2.5\ \mathrm{A}$.
• C. 2.75 · Flux density omitted
  Divides the force by the length alone and leaves the flux density out of the rearrangement: $3.08 \div 1.12 = 2.75\ \mathrm{A}$. There are three factors on the right of $F = BIL$, so making $I$ the subject divides by $1.1\ \mathrm{T}$ as well as by $1.12\ \mathrm{m}$.
• E. 2.8 · Length omitted
  Divides the force by the flux density alone: $3.08 \div 1.1 = 2.8\ \mathrm{A}$. 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 $1.12\ \mathrm{m}$ to be taken out of it.

Common Mistake (⚠️):
Using one crossing of the field as though a single conductor were in it. The wire makes 4 crossings of the field with the current the same way each time, so the conductor in the field is $1.12\ \mathrm{m}$ rather than $0.28\ \mathrm{m}$, and the current needed for the stated force is smaller than a single strand would need by a factor of 4.

Takeaway (📌):
$F = BIL$ counts metres of conductor inside the field, not wires. Wind a wire so that 4 strands cross the same gap carrying the current the same way and the length to use is $4 \times 0.28 = 1.12\ \mathrm{m}$, so the current needed for a fixed force is a factor of 4 lower than one strand would need.

Correct Answer: D. Stages 1, 4 and 6 only

Fastest Approach (🚀):
Strike out every stage in which nothing changes: the magnet at rest, the coil and magnet carried together, the current steady. What survives is 1, 4 and 6.

Step-by-Step Breakdown:

1. Stages 1 and 2: a moving magnet and a stationary one

In stage 1 the magnet moves relative to the coil, so its field lines are cut by the turns of wire and a voltage is induced: the voltmeter deflects. In stage 2 the magnet sits at rest inside the coil. The field through the coil is strong but constant, no field lines are being cut, and the voltmeter reads zero. Stage 1 yes, stage 2 no.

2. Stage 3: coil and magnet moving together

The coil is carried across the bench, but the magnet is carried with it, still at rest relative to the turns. What matters is relative motion between the wire and the field: there is none, the field through the coil does not change, and no voltage is induced. Stage 3 no.

3. Stages 4 and 5: current switched on, then steady

With the magnet gone, the only field comes from the second coil. While its current is rising, the field it produces is growing, and that changing field threads the first coil, so a voltage is induced during stage 4. Once the current is steady in stage 5, the field is steady too, nothing is changing, and the induced voltage is zero. Stage 4 yes, stage 5 no.

4. Stage 6: current switched off

Opening the switch makes the current, and with it the field through the coils, fall to zero. A falling field is a changing field, so a voltage is induced again while it collapses. Stage 6 yes.

Collecting the stages in which something changes: 1, 4 and 6.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. Stage 1 only · Changing field ignored
  Accepts only the moving magnet and misses that a changing current in the second coil changes the field through the first coil just as effectively, so stages 4 and 6 also induce a voltage.
• B. Stages 1 and 3 only · Relative motion misread
  Treats any motion of the coil as cutting field lines, so adds stage 3. The magnet travels with the coil, so there is no relative motion and no change in the field through the turns; and stages 4 and 6 are again missed.
• C. Stages 1 and 4 only · Switch-off ignored
  Includes switching on but not switching off, as if only a growing field induces a voltage. A field falling to zero is changing just as much as one rising from zero, so stage 6 also induces a voltage.
• E. Stages 4 and 6 only · Moving magnet ignored
  Counts only the electromagnet stages, as though induction needed a current-carrying coil. A permanent magnet pushed into the coil makes the turns cut its field lines and induces a voltage in stage 1 as well.
• F. Stages 1, 3, 4 and 6 only · Relative motion misread
  Adds stage 3 to the correct three, reasoning that a coil in motion must be cutting field lines. Since the magnet moves with it, the field through the coil is unchanged and nothing is induced.
• G. Stages 1, 4, 5 and 6 only · Steady field counted
  Adds stage 5, taking a steady current in the second coil to be enough. A steady current gives a steady field through the first coil, and a field that is not changing induces no voltage.
• H. Stages 1, 2, 4, 5 and 6 only · Steady field counted
  Adds stages 2 and 5 to the correct three, taking the resting magnet of stage 2 and the steady current of stage 5 to induce a voltage simply because a field is present. Both give a constant field through the coil, and a constant field induces nothing.

Common Mistake (⚠️):
Counting stage 5 because a current is flowing and a field is present. Induction needs a change, not a field: a steady current gives a steady field through the first coil, and a steady field induces nothing, exactly as the resting magnet in stage 2 induces nothing.

Takeaway (📌):
Ask of each stage one question only: is the field through the coil changing, or are the turns cutting field lines? Motion together, rest, and steady current all answer no; approaching, switching on and switching off all answer yes.

Correct Answer: D. 3.9

Fastest Approach (🚀):
Eliminate before calculating. The path is $153\ \mathrm{m}$ long and the straight line between its ends is shorter than that, so the answer has to come out below $5.1\ \mathrm{m\,s}^{-1}$, which is what the path length over $30\ \mathrm{s}$ gives. Neither leg's own average speed can be the answer either, because the average velocity is spread over the whole $30\ \mathrm{s}$.

Step-by-Step Breakdown:

1. Separate the path from the displacement

The sample case is carried $108\ \mathrm{m}$ horizontally and then falls $45\ \mathrm{m}$, so it travels $108 + 45 = 153\ \mathrm{m}$ of path. Average velocity does not use that figure. It uses the straight line from the start to the landing point, and the flight and the drop are perpendicular, so

2. Time for the whole journey

The sample case leaves the delivery drone at rest, so the fall obeys $h = \frac{1}{2}gt^{2}$ with $g = 10\ \mathrm{m\,s}^{-2}$:

The flight took $27\ \mathrm{s}$, so the clock runs for $27 + 3 = 30\ \mathrm{s}$.

3. One division

The key is $3.9\ \mathrm{m\,s}^{-1}$.

Why the Other Options Are Wrong (❌):

• A. 2.1 · Displacement found by subtraction
  The two legs are subtracted rather than combined at right angles: the difference between $108\ \mathrm{m}$ and $45\ \mathrm{m}$ is $63\ \mathrm{m}$, and $63 \div 30 = 2.1\ \mathrm{m\,s}^{-1}$. Subtracting is what a journey out and back along one line would need. These legs are perpendicular, so the straight line between the ends is $\sqrt{11664 + 2025} = 117\ \mathrm{m}$.
• B. 5.1 · Distance used instead of displacement
  The path length is used in place of the displacement: $108 + 45 = 153\ \mathrm{m}$, then $153 \div 30 = 5.1\ \mathrm{m\,s}^{-1}$. That is the average speed of the journey, which measures how far the sample case travelled rather than how far it ended up from where it started.
• C. 4 · Flight leg treated as the whole journey
  Only the horizontal flight is used: $108 \div 27 = 4\ \mathrm{m\,s}^{-1}$, the average speed of that leg on its own. It leaves out the $45\ \mathrm{m}$ the sample case then falls, and the $3\ \mathrm{s}$ the fall adds to the clock.
• E. 15 · Fall treated as the whole journey
  Only the drop is used: $45 \div 3 = 15\ \mathrm{m\,s}^{-1}$, the average speed of the fall on its own. The sample case was also carried $108\ \mathrm{m}$ horizontally over $27\ \mathrm{s}$, and both legs belong to the journey.

Common Mistake (⚠️):
Adding the two legs and dividing by the total time: $108 + 45 = 153\ \mathrm{m}$, and $153 \div 30 = 5.1\ \mathrm{m\,s}^{-1}$. That is the average speed of the journey. It measures how far the sample case travelled rather than how far it finished from where it began, and for a journey that turns a corner it is the larger of the two.

Takeaway (📌):
For velocity, draw one arrow from the first point to the last and measure only that arrow; for speed, walk the whole path. Here the arrow is the hypotenuse, $117\ \mathrm{m}$, while the walk is $153\ \mathrm{m}$, and the clock runs through both legs, $27\ \mathrm{s}$ of flight and $3\ \mathrm{s}$ of falling.

Correct Answer: B. 3.5

Fastest Approach (🚀):
Form the two totals before touching anything else, $2100\,\mathrm{m}$ over $600\,\mathrm{s}$, then divide once. The stage speeds $4$ and $7.5$ are quicker to work out and both appear on the option list, which is exactly why they are worth resisting.

Step-by-Step Breakdown:

1. Add the distances, then add the times

The journey covers $1200\,\mathrm{m}$ before the wait and $900\,\mathrm{m}$ after it, so the total distance is

The clock runs for $300\,\mathrm{s}$, then $180\,\mathrm{s}$, then $120\,\mathrm{s}$. 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 $2100\,\mathrm{m}$ by the $420\,\mathrm{s}$ actually spent moving would give $5\,\mathrm{m\,s^{-1}}$, 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 $3.5$.

Why the Other Options Are Wrong (❌):

• A. 1.5 · Counts only the distance travelled after the wait
  Takes only the distance covered after the wait, $900\,\mathrm{m}$, and divides it by the full $600\,\mathrm{s}$: $900 \div 600 = 1.5$. The $1200\,\mathrm{m}$ covered before the wait is part of the journey too, so the numerator is short by exactly that much.
• C. 7.5 · Quotes the speed of the final stage
  The last stage covers $900\,\mathrm{m}$ in $120\,\mathrm{s}$, so $900 \div 120 = 7.5$. 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.
• D. 5 · Leaves the stopped time out of the total time
  Uses the moving time only, $300 + 120 = 420\,\mathrm{s}$, giving $2100 \div 420 = 5$. 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.
• E. 4 · Quotes the speed of the first stage
  The first stage covers $1200\,\mathrm{m}$ in $300\,\mathrm{s}$, so $1200 \div 300 = 4$. That is the speed of one stage rather than an average: the journey runs on for another $180\,\mathrm{s}$ of waiting and $120\,\mathrm{s}$ of motion after it.

Common Mistake (⚠️):
Stopping the clock during the wait. Dividing the total $2100\,\mathrm{m}$ by only the $420\,\mathrm{s}$ actually spent moving gives $5\,\mathrm{m\,s^{-1}}$, which is the average speed while moving. The question asks for the average over the whole journey, and the $180\,\mathrm{s}$ spent at a landing stage 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: C. $3.6\,\mathrm{m\,s^{-2}}$

Fastest Approach (🚀):
Go straight to $a = g\left[1 - \left(\dfrac{v}{v_{t}}\right)^{2}\right]$. The ratio here is $\dfrac{4}{5}$, so the bracket is $1 - \dfrac{16}{25} = \dfrac{9}{25}$ and one multiplication gives $3.6\,\mathrm{m\,s^{-2}}$, with no drag constant and no mass written down at all. Any option as large as 10 can go on sight, because the drag is already acting.

Step-by-Step Breakdown:

1. Turn the terminal speed into a force equation

At the terminal speed the steel ball bearing is no longer speeding up, so the resultant force on it is zero and the drag has grown until it matches the weight. Writing the drag as $cv^{2}$,

Neither the mass $m$ nor the drag constant $c$ is known, and neither is needed: this one equation lets the drag at any other speed be written as a share of the weight.

2. Square the speed ratio

The speed at the instant in question is $40\,\mathrm{m\,s^{-1}}$ against a terminal speed of $50\,\mathrm{m\,s^{-1}}$, a fraction $\dfrac{4}{5}$ of it. The drag goes as the square of the speed, so it is that fraction squared, times the weight:

3. Apply Newton's second law

Taking downwards as positive, the resultant is the weight minus the drag:

The mass cancels, which is why no value for it was ever wanted, and with the acceleration of free fall taken as 10 this comes to $3.6\,\mathrm{m\,s^{-2}}$.

Sanity check: the steel ball bearing is still speeding up and is already being resisted, so the answer has to lie between zero and 10, and it does.

The key is $3.6\,\mathrm{m\,s^{-2}}$.

Why the Other Options Are Wrong (❌):

• A. $10.0\,\mathrm{m\,s^{-2}}$ · Drag ignored
  Treats the steel ball bearing as though gravity were the only force acting on it, so the acceleration is the acceleration of free fall itself, 10. At $40\,\mathrm{m\,s^{-1}}$ the drag has already grown to $\dfrac{16}{25}$ of the weight, so only $\dfrac{9}{25}$ of the weight is left to accelerate the steel ball bearing and the acceleration is below 10.
• B. $2.0\,\mathrm{m\,s^{-2}}$ · Drag taken as proportional to the speed
  Scales the drag by the speed ratio itself instead of by its square, reading a speed $\dfrac{4}{5}$ of the terminal speed as a drag $\dfrac{4}{5}$ of the weight and leaving $1 - \dfrac{4}{5} = \dfrac{1}{5}$ of it, which is $2.0\,\mathrm{m\,s^{-2}}$. The drag goes as $v^{2}$, so the ratio is squared before it is applied: the drag is $\dfrac{16}{25}$ of the weight, a smaller share than $\dfrac{4}{5}$, and the acceleration is $3.6\,\mathrm{m\,s^{-2}}$.
• D. $6.4\,\mathrm{m\,s^{-2}}$ · Drag fraction quoted as the resultant
  Squares the ratio correctly to $\dfrac{16}{25}$ and then quotes that as what is LEFT of the weight rather than as what the drag takes away, giving $6.4\,\mathrm{m\,s^{-2}}$. The squared fraction is the drag; the resultant is the rest of the weight, $\dfrac{9}{25}$ of it, and the acceleration is $3.6\,\mathrm{m\,s^{-2}}$.
• E. $18.0\,\mathrm{m\,s^{-2}}$ · Resultant fraction applied to the terminal speed
  Reaches the right fraction, $\dfrac{9}{25}$ of the weight, and then multiplies the terminal speed by it instead of the acceleration of free fall: $\dfrac{9}{25} \times 50\,\mathrm{m\,s^{-1}}$ is a speed, and its number is quoted here as an acceleration, $18.0\,\mathrm{m\,s^{-2}}$. The fraction multiplies the acceleration of free fall, 10, which gives $3.6\,\mathrm{m\,s^{-2}}$.

Common Mistake (⚠️):
Scaling the drag by the speed ratio instead of by its square. That reads a speed of $\dfrac{4}{5}$ of the terminal speed as a drag of $\dfrac{4}{5}$ of the weight and gives $2.0\,\mathrm{m\,s^{-2}}$. The stem says the drag is proportional to the square of the speed, so the share of the weight it takes is $\left(\dfrac{4}{5}\right)^{2} = \dfrac{16}{25}$, a smaller share, and the true acceleration $3.6\,\mathrm{m\,s^{-2}}$ is the larger of the two.

Takeaway (📌):
A terminal speed is a free force equation. Drag equals weight there, so at any lower speed the drag is the weight times the SQUARE of the speed ratio and $a = g\left[1 - \left(\dfrac{v}{v_{t}}\right)^{2}\right]$. Mass and drag constant never appear, and the whole question is one squaring and one subtraction.

Correct Answer: A. 3.30

Fastest Approach (🚀):
Do not work out the three forces one at a time. One line does it, $(14.40 - 3.60) \times \left(1 - \left(\tfrac{5}{6}\right)^{2}\right) = 3.30$. And $10.80$ can be struck out on sight: that is the resultant at the moment of release, and some drag is already acting at the instant asked about.

Step-by-Step Breakdown:

1. Read the terminal drag off the steady fall

Where the fall is steady the resultant force is zero, so the three forces balance and the drag is the difference between the two constant forces:

2. Scale that drag down to the instant asked about

The drag is proportional to the square of the speed and the speed there is five sixths of the terminal speed, so the drag is $\left(\tfrac{5}{6}\right)^{2}$ of its terminal value:

3. Add the three forces

The weight acts downwards, the upthrust and the drag upwards:

4. Sanity check

At release the drag is zero and the resultant is the whole $10.80\ \mathrm{N}$; at the terminal speed the resultant is zero. The body here is somewhere between those two states, so the answer has to lie between $0$ and $10.80\ \mathrm{N}$, and $3.30\ \mathrm{N}$ does.

The key is $3.30$.

Why the Other Options Are Wrong (❌):

• B. 1.80 · Wrong power of the speed in the drag law
  Scaled the terminal drag by $\left(\tfrac{5}{6}\right)$, as though the drag were proportional to the speed. That gives a drag of $9.00\ \mathrm{N}$ at five sixths of the terminal speed and a resultant of $14.40 - 3.60 - 9.00 = 1.80\ \mathrm{N}$. The stem states that the drag is proportional to the square of the speed, so the terminal drag is scaled by $\left(\tfrac{5}{6}\right)^{2}$.
• C. 7.50 · Answered the wrong quantity
  Found the drag at five sixths of the terminal speed correctly, $10.80 \times \left(\tfrac{5}{6}\right)^{2} = 7.50\ \mathrm{N}$, and stopped there. The question asks for the resultant of all three forces, which is $14.40 - 3.60 - 7.50 = 3.30\ \mathrm{N}$.
• D. 10.80 · Drag at the stated speed omitted
  Used the two constant forces only, $14.40 - 3.60 = 10.80\ \mathrm{N}$. That is the resultant at the moment of release, when the speed is zero and so is the drag. At five sixths of the terminal speed the drag has grown to $7.50\ \mathrm{N}$, and it acts on the body along with the other two.
• E. 0.80 · Terminal drag taken as the weight
  Balanced the terminal drag against the weight alone, $14.40\ \mathrm{N}$, leaving the upthrust out of that balance. The drag at five sixths of the terminal speed then comes out as $14.40 \times \left(\tfrac{5}{6}\right)^{2} = 10.00\ \mathrm{N}$, and the sum as $14.40 - 3.60 - 10.00 = 0.80\ \mathrm{N}$. In a liquid the steady fall balances the drag against the weight minus the upthrust, $10.80\ \mathrm{N}$.

Common Mistake (⚠️):
Using the wrong power of the speed. Scaling the terminal drag of $10.80\ \mathrm{N}$ by $\left(\tfrac{5}{6}\right)$ rather than $\left(\tfrac{5}{6}\right)^{2}$ treats the drag as proportional to the speed, and gives $1.80\ \mathrm{N}$ for the resultant. The power in the drag law is doing real work here: it is what fixes how much of the weight minus the upthrust is still unbalanced at a given fraction of the terminal speed.

Takeaway (📌):
The terminal drag need not be handed to you, because it can be read off the steady fall as the weight minus the upthrust. Scale it by the speed ratio raised to the power in the drag law to get the drag now, and the resultant is the part of that driving force the drag has not yet cancelled: $F = (W - U)\left(1 - \left(\frac{v}{v_T}\right)^{p}\right)$, with $p$ whatever power the stem states.

Correct Answer: B. $9.00\,\mathrm{W}$

Fastest Approach (🚀):
Collect the factors before touching the $12\,\mathrm{W}$. Here they are $2$, $2.25$, $\tfrac{1}{3}$ and $0.5$; the $2$ and the $0.5$ cancel outright, leaving $2.25/3=0.75$, so the answer is three quarters of $12$ and the multiplication is a single step.

Step-by-Step Breakdown:

Step 1: Identify the four factors that set a conduction rate

For a lagged bar carrying heat steadily along its length, the rate of transfer depends on the thermal conductivity $k$ of the material, the cross-sectional area $A$ the energy passes through, the temperature difference $\Delta T$ between the ends, and the length $L$ over which that difference falls: rate $\propto kA\Delta T/L$. Every quantity here is given as a ratio, so the value of $k$ for brass is never needed.

Step 2: Convert the diameter change into an area change

The bars are cylinders, so $A=\pi d^{2}/4$ and the area follows the square of the diameter. A diameter $1.5$ times as large gives an area factor of $1.5^{2}=2.25$.

Step 3: Collect the length and temperature-difference factors

Length sits underneath, so a bar three times as long contributes a factor $\tfrac{1}{3}$. The brass bar spans $100-20=80\,\mathrm{K}$ and the replacement spans $60-20=40\,\mathrm{K}$, giving a factor $40/80=0.5$.

Step 4: Multiply every factor onto the measured rate

Conductivity contributes a factor $2$, so the combined factor is $2\times2.25\times\tfrac{1}{3}\times0.5=0.75$. The replacement therefore conducts $0.75\times12=9.00\,\mathrm{W}$.

Sanity check: the combined factor came out as $0.75$, which is less than $1$, so the new rate must sit below the $12\,\mathrm{W}$ measured for the brass bar, and $9.00\,\mathrm{W}$ is three quarters of $12\,\mathrm{W}$.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $4.50\,\mathrm{W}$ · Omits the conductivity change
  Applies the geometry and temperature changes but treats the replacement bar as though it were brass, dropping the factor of $2$: $12\times2.25\div3\times0.5=4.50\,\mathrm{W}$. The material is one of the four factors and the alloy conducts twice as well, so that factor belongs in the product.
• C. $36.0\,\mathrm{W}$ · Inverts the temperature-difference ratio
  Takes the ratio the wrong way up, using $80/40=2$ instead of $40/80=0.5$: $12\times2\times2.25\div3\times2=36.0\,\mathrm{W}$. The replacement bar spans the smaller temperature difference, $60-20=40\,\mathrm{K}$, so this factor must reduce the rate.
• D. $6.00\,\mathrm{W}$ · Treats area as proportional to diameter
  Uses the diameter ratio $1.5$ directly as the area ratio instead of squaring it: $12\times2\times1.5\div3\times0.5=6.00\,\mathrm{W}$. For a cylinder $A=\pi d^{2}/4$, so a diameter $1.5$ times as large gives $2.25$ times the area.
• E. $13.5\,\mathrm{W}$ · Cubes the diameter ratio
  Cubes the diameter ratio instead of squaring it, using $1.5^{3}=3.375$: $12\times2\times3.375\div3\times0.5=13.5\,\mathrm{W}$. Conduction depends on the area the energy crosses, and that area carries the diameter squared, so the diameter ratio is squared and not cubed.
• F. $2.25\,\mathrm{W}$ · Inverts the conductivity ratio
  Reads the alloy as the poorer conductor and multiplies by $0.5$ instead of $2$: $12\times0.5\times2.25\div3\times0.5=2.25\,\mathrm{W}$. The stem states that the alloy's conductivity is twice that of brass, so this factor raises the rate.
• G. $81.0\,\mathrm{W}$ · Multiplies by the length ratio instead of dividing
  Puts length on the wrong side of the relation, so the extra length is treated as helping the flow: $12\times2\times2.25\times3\times0.5=81.0\,\mathrm{W}$. A longer bar spreads the same temperature difference over a greater distance, so the length ratio must divide.
• H. $18.0\,\mathrm{W}$ · Ignores the change in temperature difference
  Applies conductivity, area and length but keeps the original $80\,\mathrm{K}$ span across the replacement bar: $12\times2\times2.25\div3=18.0\,\mathrm{W}$. The replacement bar's ends sit at $60\,^{\circ}\mathrm{C}$ and $20\,^{\circ}\mathrm{C}$, a difference of only $40\,\mathrm{K}$, which halves the rate.

Common Mistake (⚠️):
Scaling the conduction rate with the diameter itself rather than with the cross-sectional area. The $1.5$ is then used once instead of twice, so the area factor becomes $1.5$ rather than $1.5^{2}=2.25$ and the answer comes out as $6.00\,\mathrm{W}$. Conduction responds to the area the energy crosses, and for a circular bar that area carries the diameter squared.

Takeaway (📌):
When a conduction question hands you a measured rate and then changes the bar, do not rebuild the whole calculation. Write the rate as proportional to $kA\Delta T/L$, turn each change into a multiplying factor, remembering that a diameter ratio must be squared before it becomes an area ratio and that a length ratio goes underneath, then multiply the factors onto the rate you were given.

Correct Answer: C. 120

Fastest Approach (🚀):
Do the period first, $T = \lambda / v = 0.5\,\mathrm{s}$, so the $3\,\mathrm{s}$ holds $6$ cycles and the answer is $6$ lots of $4A$. Any option that is a distance the wave covered, $40$ for one cycle or $240$ for the whole time, is answering a different question.

Step-by-Step Breakdown:

1. Turn the wavelength and the speed into a period

The pattern advances one whole wavelength in one period, so $T = \frac{\lambda}{v} = \frac{40}{80} = 0.5\,\mathrm{s}$.

2. Count the complete cycles in the time given

$\frac{3}{0.5} = 6$, so the $3\,\mathrm{s}$ holds $6$ complete cycles.

3. One cycle is four amplitudes of path

The wave is transverse: every point of the cord moves across the direction of travel, and no part of the cord travels along with the wave. Whichever part of its swing a point starts from, one complete cycle takes it out to one extreme, across to the other and back to where it began, a path of $4A = 4 \times 5 = 20\,\mathrm{cm}$.

4. Scale to the whole time

$20 \times 6 = 120\,\mathrm{cm}$.

The wave pattern meanwhile advances $vt = 80 \times 3 = 240\,\mathrm{cm}$ along the cord, which is the pattern moving and not the cord.

The key is $120$.

Why the Other Options Are Wrong (❌):

• A. 20 · Path in one cycle only, not scaled to the time
  $4A = 4 \times 5 = 20\,\mathrm{cm}$ is the path a point covers in ONE cycle. The $3\,\mathrm{s}$ holds $6$ complete cycles, so this figure has still to be multiplied by $6$.
• B. 240 · The wave's travel given instead of the medium's path
  $vt = 80 \times 3 = 240\,\mathrm{cm}$ is how far the wave pattern advances along the cord. In a transverse wave the cord does not move along its own length at all; only the pattern does, and a point of the cord moves across it.
• D. 60 · Per-cycle path multiplied by the time in seconds
  Uses the correct $4A = 20\,\mathrm{cm}$ for one cycle but multiplies it by the $3$ seconds instead of by the number of cycles: $20 \times 3 = 60\,\mathrm{cm}$. One cycle lasts $0.5\,\mathrm{s}$, so the $3\,\mathrm{s}$ holds $6$ cycles, not $3$.
• E. 40 · One wavelength quoted as the path of one point
  Treats a point of the cord as moving forward with the wave, one wavelength of $40\,\mathrm{cm}$ for each cycle, and then quotes a single cycle. Both halves are wrong: the point moves across the cord rather than along it, and the $3\,\mathrm{s}$ holds $6$ cycles.

Common Mistake (⚠️):
Giving how far the wave travels instead of how far the cord moves: $vt = 80 \times 3 = 240\,\mathrm{cm}$. In a transverse wave nothing is carried along the cord except the pattern itself; each point of the cord moves across it and is back where it started at the end of every cycle.

Takeaway (📌):
One complete cycle of any oscillation is four amplitudes of path, so the answer is $4A$ times the number of cycles. Take the number of cycles from $t \div T$ with $T = \lambda / v$, never from the time in seconds on its own, and remember that $vt$ is where the pattern got to, not where the medium went.

Correct Answer: C. 8

Fastest Approach (🚀):
The lowest-to-highest height $0.2\,\mathrm{m}$ is half a cycle of path, so a whole cycle is twice it, $0.4\,\mathrm{m}$. Multiply by the $20$ cycles and stop. Any option built from the $2\,\mathrm{m}$ crest spacing is answering a different question, because that spacing cannot affect how far the tape moves.

Step-by-Step Breakdown:

1. Count the cycles

A crest reaches the tape every $2\,\mathrm{s}$, so that is the period of its motion. In $40\,\mathrm{s}$ it therefore completes

complete cycles.

2. Decide what the tape actually does

The wave carries energy along, but it does not carry the spring along with it: each part of the spring, and anything riding on it, moves about one fixed place and comes back to where it started at the end of every cycle. So the tape does not travel with the crests, and the $2\,\mathrm{m}$ between them never enters the answer. The path to measure is the up-and-down one.

3. Measure one cycle of that path

In one cycle the tape goes up $0.2\,\mathrm{m}$ from its lowest position to its highest and comes back down again, a path of

4. Multiply

Check: sharing $8\,\mathrm{m}$ back among the $20$ cycles returns $0.4\,\mathrm{m}$ each, which is the rise of $0.2\,\mathrm{m}$ plus the equal fall.

The key is $8$.

Why the Other Options Are Wrong (❌):

• A. 40 · The object taken to travel along with the wave
  Treats the tape as moving along with the crests. The wave profile advances $2\,\mathrm{m}$ every $2\,\mathrm{s}$, so over $40\,\mathrm{s}$ it covers $40\,\mathrm{m}$. That is the progress of the wave, not of the spring: a wave moves energy along and leaves the matter oscillating where it was.
• B. 4 · Only the rise counted, not the fall
  Counts the $0.2\,\mathrm{m}$ rise in each cycle and forgets the equal fall, giving $20 \times 0.2 = 4\,\mathrm{m}$. A full cycle brings the tape back to where it started, so the path it covers in that cycle is $0.4\,\mathrm{m}$, twice this figure.
• D. 16 · Wave speed used where the period belongs
  Works out the wave's speed from $2\,\mathrm{m}$ in $2\,\mathrm{s}$, divides the $40\,\mathrm{s}$ by that instead of by the period, then multiplies by the $0.4\,\mathrm{m}$ path of one cycle to reach $16\,\mathrm{m}$. The number of cycles is a count of how many crests arrive, so it is the period in seconds that divides the time, never a speed.
• E. 2 · Crest spacing quoted as the distance travelled
  Quotes the $2\,\mathrm{m}$ between one crest and the next. That is a spacing in the wave pattern, not a distance anything travels: the tape never moves from one crest to the next, and a distance travelled would have to depend on the $40\,\mathrm{s}$ of watching, which this figure does not.

Common Mistake (⚠️):
Counting the rise and forgetting the fall. Each cycle takes the tape up $0.2\,\mathrm{m}$ and then down the same $0.2\,\mathrm{m}$, so one cycle is $0.4\,\mathrm{m}$ of path, not $0.2\,\mathrm{m}$. Using the rise alone gives $4\,\mathrm{m}$, which is exactly half of the distance asked for.

Takeaway (📌):
For anything riding on a wave, count the cycles and multiply by the length of one up-and-down path, which is twice the lowest-to-highest height. The crest spacing and the wave's own speed never appear in that answer.