ESAT Practice Set 1A Physics Worked Solutions

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 (📌):
The Cheat Code: 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: 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 (📌):
The Cheat Code: 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 (📌):
The Cheat Code: 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.

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 (📌):
The Cheat Code: 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: 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 marked instant

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 the marked instant 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 the marked instant 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 marked instant 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 the marked instant 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 Cheat Code: the terminal drag is never handed to you, it is 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.