ESAT Practice Set 1B Physics Worked Solutions

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 bends lie 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 folding is counted 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 (📌):
The Cheat Code: $F = BIL$ counts metres of conductor inside the field, not wires. Fold 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: 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 (📌):
The Cheat Code: 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 · 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 (📌):
The Cheat Code: 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: 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 in the iron core 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 in the core, 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 cuts 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
  Counts every stage in which a magnet or a current is present, including the resting magnet of stage 2 and the steady current of stage 5. 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 (📌):
The Cheat Code: 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: 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
  Scales with volume rather than with the cross-section, 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, not on how much metal the bar contains, so the diameter 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 (📌):
The Cheat Code: 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.