ESAT Physics Mock Module 1 Worked Solutions

Correct Answer: D. Acceleration

Fastest Approach (🚀):
Speed, mass, energy and temperature need no direction.
Acceleration does.

Matches Option D.

Step-by-Step Breakdown:

1. The test

A scalar is fully described by a magnitude and a unit. A vector also needs a direction.

2. Apply it

Speed: $30\ \text{m/s}$ is complete. Scalar. (Its vector partner is velocity.)
Mass: $5\ \text{kg}$ is complete. Scalar. (Its vector partner is weight, a force.)
Energy: $200\ \text{J}$ is complete. Scalar.
Temperature: $20\ ^{\circ}\text{C}$ is complete. Scalar.
Acceleration: $2\ \text{m/s}^{2}$ is not complete — you need to know which way. Vector.

3. The pairs worth knowing

distance (scalar) and displacement (vector)
speed (scalar) and velocity (vector)
mass (scalar) and weight (vector)

Each pair is a favourite of examiners precisely because the words are used loosely in ordinary English.

4. Why it matters in a calculation

Vectors need directions assigning before you add them. A car slowing down has a negative acceleration relative to its motion, and a ball thrown up has constant downward acceleration even at the instant its velocity is zero.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. Speed — Scalar
  Scalar — velocity is its vector counterpart.
• B. Mass — Scalar
  Scalar — weight is the associated vector.
• C. Energy — Scalar
  Scalar — energy has magnitude only.
• E. Temperature — Scalar
  Scalar.

Common Mistake (⚠️):
Calling energy a vector because it can be transferred in a direction. Energy has no direction of its own; only the transfer process does.

Takeaway (📌):
Scalar if a number and unit say everything; vector if a direction is still missing. Learn the three classic pairs.

Correct Answer: B. $2.0\times 10^{10}$

Fastest Approach (🚀):
$\dfrac{3.2}{1.6} = 2$ and $10^{-9-(-19)} = 10^{10}$.
$n = 2.0\times 10^{10}$.

Matches Option B.

Step-by-Step Breakdown:

1. What charging by friction actually moves

Rubbing does not create charge. It transfers electrons from one surface to the other: the polythene gains electrons and becomes negative, and the cloth is left equally positive. Protons never move — they are bound in the nuclei.

2. Divide the total by the individual

$n = \dfrac{\text{total charge}}{\text{charge per electron}} = \dfrac{3.2\times 10^{-9}}{1.6\times 10^{-19}}$

3. Split it into mantissa and index

$\dfrac{3.2}{1.6} = 2$ and $\dfrac{10^{-9}}{10^{-19}} = 10^{-9+19} = 10^{10}$

$n = 2.0\times 10^{10}$ electrons.

4. Check the size is sensible

Twenty billion electrons sounds enormous, but the charge on each is so tiny that the rod's total charge is only a few nanocoulombs. Any answer smaller than one would be impossible — you cannot transfer a fraction of an electron.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $5.0\times 10^{-11}$ — Inverted Division
  Dividing the electron charge by the total charge.
• C. $2.0\times 10^{9}$ — Index Error
  Subtracting the indices as $-9-19+19$ or losing a power of ten.
• D. $5.1\times 10^{-28}$ — Operation Error
  Multiplying the two charges instead of dividing.
• E. $2.0\times 10^{11}$ — Index Error
  Using $10^{11}$, off by one power of ten.

Common Mistake (⚠️):
Dividing the other way round, giving $5.0\times 10^{-11}$. A count of objects must come out greater than one, which rules that answer out before any arithmetic is checked.

Takeaway (📌):
Number of carriers is always total charge over charge per carrier. Subtract the indices; do not try to do it on a calculator you are not allowed.

Correct Answer: E. $252\,000\ \text{J}$

Fastest Approach (🚀):
$2\times 4200 = 8400$, then $8400\times 30 = 252\,000\ \text{J}$.

Matches Option E.

Step-by-Step Breakdown:

1. Apply the equation

$E = mc\Delta\theta = 2\times 4200\times 30$

2. Multiply in a convenient order

$2\times 4200 = 8400$
$8400\times 30 = 252\,000\ \text{J}$

That is $252\ \text{kJ}$ — roughly what a $1\ \text{kW}$ kettle delivers in four minutes.

3. Why a temperature difference needs no conversion

$\Delta\theta$ is a change, and a change of $30\ ^{\circ}\text{C}$ is a change of $30\ \text{K}$. Only absolute temperatures need converting; differences do not.

4. What makes water unusual

$4200\ \text{J/kg}^{\circ}\text{C}$ is very high — several times that of most metals. That is why water is used as a coolant and in central heating, and why coastal climates are milder: a large mass of water absorbs a lot of energy for a small temperature rise.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $8400\ \text{J}$ — Factor Omitted
  Computing $mc$ and omitting the temperature change.
• B. $2520\ \text{J}$ — Decimal Slip
  A factor of $100$ slip.
• C. $126\,000\ \text{J}$ — Factor Omitted
  Using $m = 1\ \text{kg}$.
• D. $63\,000\ \text{J}$ — Arithmetic Error
  Using $\Delta\theta = 15$, or halving twice.

Common Mistake (⚠️):
Omitting a factor — usually the mass or the temperature change — which lands on $126\,000$ or $8400$. Both are in the options.

Takeaway (📌):
$E = mc\Delta\theta$ for a temperature change. A temperature difference in $^{\circ}\text{C}$ equals the same difference in kelvin, so no conversion is needed.

Correct Answer: A. Speed unchanged; pressure doubles

Fastest Approach (🚀):
Temperature fixed $\Rightarrow$ speed fixed.
Half the volume $\Rightarrow$ twice the collision rate $\Rightarrow$ double pressure.

Matches Option A.

Step-by-Step Breakdown:

1. What sets the particle speed

The average kinetic energy of the particles depends only on the temperature. The temperature is held constant, so the average speed is unchanged.

2. What sets the pressure

Pressure comes from particles colliding with the container walls. Halving the volume means each particle has half the distance to travel between wall collisions, so collisions happen twice as often — with the same force per collision, since the speed has not changed.

Twice the collision rate over the same wall area gives twice the pressure.

3. The quantitative statement

$pV = \text{constant}$ at constant temperature, so halving $V$ doubles $p$.

4. Contrast with heating

Heat the gas at constant volume instead and the particles speed up, striking the walls both harder and more often — so pressure rises for a different reason. Keeping the two mechanisms separate is what this question is testing.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. Speed doubles; pressure doubles — Speed Changed
  Speed depends on temperature alone, which is unchanged.
• C. Speed unchanged; pressure halves — Direction Reversed
  Pressure rises when a gas is compressed, not falls.
• D. Speed halves; pressure doubles — Speed Changed
  Compression does not slow the particles down.
• E. Speed doubles; pressure unchanged — Pressure Unchanged
  Halving the volume must raise the pressure.

Common Mistake (⚠️):
Assuming compression speeds the particles up. Pushing the walls in does work on the gas and would warm it, but the question specifies constant temperature, so that energy is removed.

Takeaway (📌):
Temperature governs particle speed; volume governs collision frequency. $pV$ is constant when the temperature is.

Correct Answer: B. $3\ \text{s}$

Fastest Approach (🚀):
$45 = \tfrac12(10)t^{2} = 5t^{2}$.
$t^{2} = 9 \implies t = 3\ \text{s}$.

Matches Option B.

Step-by-Step Breakdown:

1. List the five quantities

$s = 45\ \text{m}$, $u = 0$ (dropped from rest), $v = ?$ (not wanted), $a = 10\ \text{m/s}^{2}$, $t = ?$ (wanted).

2. Choose the equation without v

$s = ut+\tfrac12 at^{2}$

3. Substitute

$45 = 0+\tfrac12(10)t^{2} = 5t^{2}$
$t^{2} = 9$
$t = 3\ \text{s}$

Take the positive root: negative time has no meaning here.

4. Check by another route

$v^{2} = u^{2}+2as = 0+2(10)(45) = 900$, so $v = 30\ \text{m/s}$. Then $t = \dfrac{v-u}{a} = \dfrac{30}{10} = 3\ \text{s}$. ✓

Note that $30$ appears among the options — it is the final speed, not the time, and is exactly what a careless reader picks up.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $4.5\ \text{s}$ — Acceleration Ignored
  Dividing $45$ by $10$, treating the motion as constant speed.
• C. $9\ \text{s}$ — Root Not Taken
  Giving $t^{2}$ rather than $t$.
• D. $1.5\ \text{s}$ — Factor Error
  Dividing the correct answer by two, or misusing $\tfrac12$.
• E. $30\ \text{s}$ — Wrong Quantity
  Giving the final speed in $\text{m/s}$, not the time.

Common Mistake (⚠️):
Using $s = vt$ with $v = 10$, giving $4.5\ \text{s}$. The object is accelerating, so it has no single speed to divide by.

Takeaway (📌):
Write out $s$, $u$, $v$, $a$, $t$ and mark the one you neither have nor want. The equation that omits it is the one to use.

Correct Answer: E. $6\ \Omega$

Fastest Approach (🚀):
Product over sum: $\dfrac{6\times 3}{9} = 2\ \Omega$.
$2+4 = 6\ \Omega$.

Matches Option E.

Step-by-Step Breakdown:

1. Deal with the parallel pair

$\dfrac{1}{R} = \dfrac{1}{6}+\dfrac{1}{3} = \dfrac{1}{6}+\dfrac{2}{6} = \dfrac{3}{6} = \dfrac{1}{2}$

so $R = 2\ \Omega$. For exactly two resistors the product-over-sum shortcut is quicker: $\dfrac{6\times 3}{6+3} = \dfrac{18}{9} = 2\ \Omega$.

2. Add the series resistor

Series resistances add directly:
$R_{\text{total}} = 2+4 = 6\ \Omega$

3. The check that catches the usual error

A parallel combination is always smaller than the smallest resistor in it. Here $2\ \Omega$ is less than $3\ \Omega$ — correct. If a parallel result ever comes out larger than one of its branches, the reciprocals have been mishandled.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $13\ \Omega$ — Parallel Ignored
  Adding all three as though the whole network were in series.
• B. $9\ \Omega$ — Parallel Ignored
  Adding $6+3$ for the parallel pair.
• C. $2\ \Omega$ — Incomplete
  Giving the parallel combination and forgetting the series resistor.
• D. $4.5\ \Omega$ — Formula Misuse
  Averaging, or using $\dfrac{6+3}{2}$.

Common Mistake (⚠️):
Adding all three resistances to get $13\ \Omega$. Only series resistances add; parallel ones combine as reciprocals, and adding them is the single most common circuit error.

Takeaway (📌):
Reduce parallel sections first, then add in series. Product over sum works for exactly two resistors and is much faster than reciprocals.

Correct Answer: C. $167\,000\ \text{J}$

Fastest Approach (🚀):
$E = mL = 0.5\times 334\,000 = 167\,000\ \text{J}$.

Matches Option C.

Step-by-Step Breakdown:

1. Which equation applies

Two different processes, two different equations:
temperature change: $E = mc\Delta\theta$
change of state: $E = mL$

The ice is already at its melting point, so only the state change happens.

2. Substitute

$E = mL = 0.5\times 334\,000 = 167\,000\ \text{J}$

3. Why the temperature does not rise

During melting, the energy goes into breaking the bonds holding the lattice together, not into speeding the particles up. The temperature stays at $0\ ^{\circ}\text{C}$ until the last of the ice has melted — which is why a drink with ice in it stays cold rather than warming steadily.

4. If the ice had started colder

Ice at $-10\ ^{\circ}\text{C}$ would need two stages: $mc\Delta\theta$ to warm it to $0\ ^{\circ}\text{C}$, then $mL$ to melt it. Watch for that phrasing — it is the standard extension of this question.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $668\,000\ \text{J}$ — Mass Inverted
  Multiplying by $2$ instead of $0.5$.
• B. $334\,000\ \text{J}$ — Mass Ignored
  Using $1\ \text{kg}$ instead of $0.5\ \text{kg}$.
• D. $0\ \text{J}$ — Wrong Equation
  Concluding that no energy is needed because $\Delta\theta = 0$.
• E. $83\,500\ \text{J}$ — Arithmetic Error
  Quartering rather than halving.

Common Mistake (⚠️):
Trying to use $mc\Delta\theta$ with $\Delta\theta = 0$ and concluding no energy is needed. Melting takes a great deal of energy at constant temperature.

Takeaway (📌):
$E = mL$ for a change of state, $E = mc\Delta\theta$ for a change of temperature. Never the same equation, and sometimes both stages in sequence.

Correct Answer: C. $50\ \text{m}$

Fastest Approach (🚀):
$s = \dfrac{u^{2}}{2a} = \dfrac{400}{8} = 50\ \text{m}$.

Matches Option C.

Step-by-Step Breakdown:

1. Set up

$u = 20\ \text{m/s}$, $v = 0$ (it stops), $a = -4\ \text{m/s}^{2}$, $s = ?$, and $t$ is not involved.

2. Use the equation without t

$v^{2} = u^{2}+2as$
$0 = 400+2(-4)s$
$8s = 400$
$s = 50\ \text{m}$

3. Handle the sign properly

Taking the direction of travel as positive makes the deceleration negative. Alternatively drop the signs entirely and write $s = \dfrac{u^{2}}{2a} = \dfrac{400}{8}$, which is safer under pressure.

4. Why braking distance grows so fast

$s \propto u^{2}$. Doubling the speed quadruples the stopping distance: at $40\ \text{m/s}$ the same car needs $200\ \text{m}$. That square relationship is the physics behind every speed limit.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $5\ \text{m}$ — Wrong Quantity
  Computing $\dfrac{u}{a}$, which is the stopping time.
• B. $100\ \text{m}$ — Factor Omitted
  Omitting the factor of $2$.
• D. $80\ \text{m}$ — Formula Misuse
  Using $u\times a$ or another mis-formed product.
• E. $25\ \text{m}$ — Factor Error
  Halving the correct answer.

Common Mistake (⚠️):
Computing $\dfrac{u}{a} = 5$ and quoting it as a distance. That is the stopping time in seconds.

Takeaway (📌):
$v^{2} = u^{2}+2as$ whenever time is absent. Stopping distance goes as the square of the speed.

Correct Answer: C. $8\ \text{V}$

Fastest Approach (🚀):
Total $12\ \Omega$; the $8\ \Omega$ takes $\tfrac{8}{12}$ of $12\ \text{V}$.
$8\ \text{V}$.

Matches Option C.

Step-by-Step Breakdown:

1. The same current flows through both

In series there is only one path, so the current is identical in both resistors:
$I = \dfrac{V}{R_{\text{total}}} = \dfrac{12}{4+8} = 1\ \text{A}$

2. Apply V = IR to the one you want

$V_8 = IR = 1\times 8 = 8\ \text{V}$

3. The ratio shortcut

Because the current is common, $V \propto R$. The $8\ \Omega$ resistor is $\dfrac{8}{12} = \dfrac23$ of the total resistance, so it takes $\dfrac23$ of the supply voltage:
$\dfrac23\times 12 = 8\ \text{V}$

4. Check they sum to the supply

$V_4 = \dfrac13\times 12 = 4\ \text{V}$, and $4+8 = 12\ \text{V}$. ✓ The two potential differences must add to the supply voltage in a series loop, which is the fastest check available.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $4\ \text{V}$ — Wrong Component
  Giving the voltage across the $4\ \Omega$ resistor instead.
• B. $12\ \text{V}$ — Series/Parallel Confusion
  Assuming the full supply appears across one resistor, as it would in parallel.
• D. $6\ \text{V}$ — Ratio Ignored
  Halving the supply, as though the resistors were equal.
• E. $1.5\ \text{V}$ — Wrong Quantity
  Computing $\dfrac{12}{8}$, which is a current, not a voltage.

Common Mistake (⚠️):
Giving $4\ \text{V}$ — the voltage across the other resistor. The larger resistance takes the larger share, not the smaller.

Takeaway (📌):
Series: same current, voltage splits in the ratio of the resistances. Parallel: same voltage, current splits in the inverse ratio.

Correct Answer: B. Convection, because warmed air expands, becomes less dense and rises

Fastest Approach (🚀):
Air is a fluid and a poor conductor.
Warm air expands $\Rightarrow$ less dense $\Rightarrow$ rises.

Matches Option B.

Step-by-Step Breakdown:

1. Rule out conduction

Conduction needs particles in fixed contact passing vibrations along. Gases have widely spaced particles, so air is a poor conductor — which is exactly why the air trapped in double glazing, foam and wool insulates so well.

2. Rule out radiation as the main route

The radiator does emit infrared, and that warms objects in line of sight. But air is largely transparent to infrared and absorbs little of it, so radiation is not what warms the bulk of the air.

3. Convection

Air in contact with the radiator is heated and expands. The same mass now occupies a greater volume, so its density falls, and the surrounding cooler, denser air sinks and displaces it upwards. The warm air spreads across the ceiling, cools, sinks at the far wall and returns — a convection current that circulates the whole room.

4. Why the causal chain matters

Option E has the right mechanism and the wrong physics: warmed fluid becomes less dense, not more. Getting the direction wrong would predict warm air pooling on the floor, which is the opposite of what happens.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. Conduction, because air particles collide and pass energy along — Wrong Mechanism
  Conduction is negligible in a gas; the particles are too far apart.
• C. Radiation, because air is an excellent absorber of infrared — Wrong Mechanism
  Air absorbs infrared poorly, which is why it is largely transparent.
• D. Conduction, because air is a good thermal conductor — Wrong Property
  Air is a very poor conductor — that is why it is used as an insulator.
• E. Convection, because warmed air becomes denser and sinks — Density Reversed
  Right mechanism, but warmed fluid becomes less dense, not more.

Common Mistake (⚠️):
Saying warm air rises 'because it is hot'. The step that matters is expansion, then reduced density, then upthrust from the denser fluid around it.

Takeaway (📌):
Solids conduct, fluids convect, and radiation alone crosses a vacuum. In convection the chain is always heat, expand, less dense, rise.

Correct Answer: A. $4\ \text{m/s}^{2}$

Fastest Approach (🚀):
$20-8 = 12\ \text{N}$.
$a = \dfrac{12}{3} = 4\ \text{m/s}^{2}$.

Matches Option A.

Step-by-Step Breakdown:

1. Find the resultant force

The two horizontal forces oppose each other:
$F_{\text{net}} = 20-8 = 12\ \text{N}$ in the direction of the pull.

The weight and the normal contact force act vertically and cancel, so they do not enter a horizontal calculation.

2. Apply Newton's second law

$a = \dfrac{F_{\text{net}}}{m} = \dfrac{12}{3} = 4\ \text{m/s}^{2}$

3. Read the result physically

The resultant is non-zero, so the trolley speeds up. Were friction also $20\ \text{N}$, the resultant would be zero and the trolley would move at constant velocity — not stop. That is Newton's first law, and it is the point most often misunderstood.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $6.7\ \text{m/s}^{2}$ — Friction Ignored
  Using the $20\ \text{N}$ pull without subtracting friction.
• C. $9.3\ \text{m/s}^{2}$ — Sign Error
  Adding the two forces to get $28\ \text{N}$.
• D. $2.7\ \text{m/s}^{2}$ — Wrong Force
  Using the friction force alone.
• E. $12\ \text{m/s}^{2}$ — Wrong Quantity
  Giving the resultant force in newtons, not the acceleration.

Common Mistake (⚠️):
Using the $20\ \text{N}$ pull alone, giving $6.7\ \text{m/s}^{2}$. Newton's second law takes the resultant of every force, not the one the question mentions first.

Takeaway (📌):
Resultant force over total mass, always in that order. Vertical forces cancel on a horizontal surface and can be ignored.

Correct Answer: D. $36\ \text{W}$

Fastest Approach (🚀):
$P = I^{2}R = 9\times 4 = 36\ \text{W}$.

Matches Option D.

Step-by-Step Breakdown:

1. List the three forms

$P = VI$, $P = I^{2}R$ and $P = \dfrac{V^{2}}{R}$ — all the same relationship with $V = IR$ substituted in different places.

2. Match to what you were given

You have $I$ and $R$, so $P = I^{2}R$ uses both directly with nothing to find first.

$P = 3^{2}\times 4 = 9\times 4 = 36\ \text{W}$

3. Confirm the long way

$V = IR = 3\times 4 = 12\ \text{V}$, then $P = VI = 12\times 3 = 36\ \text{W}$. ✓ Same answer, one extra step, one extra chance to slip.

4. Why the current is squared

Doubling the current doubles the voltage across the resistor as well, so the power goes up by a factor of four. That is why transmission cables carry high voltage and low current: halving the current quarters the heat wasted in the wires.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $12\ \text{W}$ — Wrong Quantity
  Computing $IR$, which gives the voltage.
• B. $0.75\ \text{W}$ — Operation Error
  Computing $\dfrac{3}{4}$.
• C. $48\ \text{W}$ — Squared Wrong Term
  Using $I\times R^{2}$ instead of $I^{2}R$.
• E. $144\ \text{W}$ — Formula Misuse
  Using $V^{2}/R$ with $V = 12$ but dividing incorrectly, or $12\times 12$.

Common Mistake (⚠️):
Computing $P = VI$ with the resistance in place of the voltage, giving $12\ \text{W}$. That is the voltage across the resistor, not its power.

Takeaway (📌):
$P = VI$, $P = I^{2}R$, $P = \dfrac{V^{2}}{R}$. Choose by which two quantities you already hold, and never square the wrong one.

Correct Answer: E. Matt black

Fastest Approach (🚀):
Best absorber = best emitter = matt black.
Fastest emission $\Rightarrow$ fastest cooling.

Matches Option E.

Step-by-Step Breakdown:

1. The governing principle

A surface that absorbs infrared well also emits it well. Matt black surfaces are the best at both; shiny silver surfaces are the worst at both, reflecting rather than absorbing and emitting poorly.

2. Apply it to cooling

The cans start hotter than the room, so they are net emitters. The best emitter loses energy fastest, so the matt black can cools quickest and the shiny silver one slowest.

3. Why texture as well as colour

Matt surfaces emit more than gloss ones of the same colour, because the roughness presents more surface area and scatters rather than reflects. Colour and texture both matter, which is why 'matt black' is specified as a pair.

4. Where this is used

Radiators in cars are painted matt black to lose heat quickly. Vacuum flasks are silvered to reflect radiation back and evacuated to remove conduction and convection — attacking all three mechanisms at once.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. Shiny silver — Reversed
  The poorest emitter, so this can cools slowest.
• B. All four cool at the same rate — Effect Denied
  Surface properties change the rate of radiation substantially.
• C. Shiny white — Reversed
  A poor emitter, and shiny surfaces reflect rather than emit.
• D. Matt white — Partially Right
  Better than shiny white but poorer than matt black.

Common Mistake (⚠️):
Assuming black absorbs but does not emit. The property is symmetric: a good absorber is necessarily a good emitter, and reversing that reverses the answer.

Takeaway (📌):
Matt black: best absorber and best emitter. Shiny silver: worst at both. Whether it heats or cools fastest depends only on which way the energy is flowing.

Correct Answer: C. $5\ \text{N}$

Fastest Approach (🚀):
A $3$-$4$-$5$ triangle.
Resultant $= 5\ \text{N}$.

Matches Option C.

Step-by-Step Breakdown:

1. Forces add as vectors

Drawing the two forces head to tail gives a right-angled triangle whose hypotenuse is the resultant.

2. Apply Pythagoras

$R = \sqrt{3^{2}+4^{2}} = \sqrt{9+16} = \sqrt{25} = 5\ \text{N}$

3. Give the direction too

A resultant force is a vector, so a complete answer states the direction as well:
$\tan\theta = \dfrac{4}{3} \implies \theta \approx 53^{\circ}$ north of east.

4. The bounds worth knowing

For two forces of $3\ \text{N}$ and $4\ \text{N}$ the resultant can be anything from $1\ \text{N}$ (opposite directions) to $7\ \text{N}$ (same direction). Both appear among the options, and both are what you get by ignoring the geometry.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $7\ \text{N}$ — Vectors Ignored
  Adding the magnitudes, correct only for parallel forces.
• B. $1\ \text{N}$ — Vectors Ignored
  Subtracting, correct only for opposing forces.
• D. $12\ \text{N}$ — Operation Error
  Multiplying the two magnitudes.
• E. $3.5\ \text{N}$ — Operation Error
  Averaging the two forces.

Common Mistake (⚠️):
Adding the magnitudes to get $7\ \text{N}$. That is only correct when the forces point the same way.

Takeaway (📌):
Perpendicular vectors combine by Pythagoras. The resultant of two forces always lies between their difference and their sum.

Correct Answer: A. $\pounds 1.44$

Fastest Approach (🚀):
$2\ \text{kW}\times 3\ \text{h} = 6\ \text{kWh}$.
$6\times 24 = 144\ \text{p} = \pounds 1.44$.

Matches Option A.

Step-by-Step Breakdown:

1. Energy in kilowatt-hours

$\text{energy} = \text{power (kW)}\times\text{time (h)} = 2\times 3 = 6\ \text{kWh}$

2. Multiply by the unit price

$6\times 24 = 144\ \text{p} = \pounds 1.44$

3. Why not joules

In SI units the energy is $2000\ \text{W}\times 10\,800\ \text{s} = 2.16\times 10^{7}\ \text{J}$ — correct, but useless for the price, which is quoted per kilowatt-hour. The kilowatt-hour exists precisely so that domestic bills avoid numbers like that.

4. Keep track of pounds and pence

The rate is in pence, so the product is in pence. Answering $\pounds 144$ or $14.4\ \text{p}$ is a decimal slip rather than a physics error, and both appear among the options.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $\pounds 0.48$ — Time Ignored
  Using $2\ \text{kWh}$ and forgetting the three hours.
• C. $\pounds 14.40$ — Decimal Slip
  A factor of ten out in the pence-to-pounds conversion.
• D. $\pounds 0.72$ — Arithmetic Error
  Using $3\ \text{kWh}$ or halving the power.
• E. $\pounds 8.64$ — Unit Error
  Using $36\ \text{kWh}$ from a units mix-up.

Common Mistake (⚠️):
Converting to joules and seconds, then multiplying by a price quoted per kilowatt-hour. The units then disagree by a factor of $3.6$ million.

Takeaway (📌):
$\text{kWh} = \text{kW}\times\text{h}$. Never convert to joules for a cost question, and check whether the answer is wanted in pence or pounds.

Correct Answer: E. The book pulling up on the Earth

Fastest Approach (🚀):
Pair = swap the two bodies, keep the force type.
Earth on book $\Rightarrow$ book on Earth.

Matches Option E.

Step-by-Step Breakdown:

1. State the rule precisely

If body A exerts a force on body B, then B exerts an equal and opposite force on A of the same type. The two forces always act on different bodies, so they can never cancel each other.

2. Build the partner

The stated force is: Earth pulls book down, gravitationally.
Swap the bodies, keep the type: book pulls Earth up, gravitationally.

3. Why Option A is the classic trap

The table pushing up on the book is also equal and opposite to the book's weight — but only because the book happens to be in equilibrium. It is a contact force, not a gravitational one, and it acts on the same body as the weight. Put the book in a lift accelerating upwards and the two are no longer equal, while the true third-law pair remains equal always.

4. The test that settles it

Ask: do the two forces act on the same object? If yes, they are not a third-law pair. Weight and normal contact force both act on the book, so they are a first-law balance, not a third-law pair.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. The table pushing up on the book — Same Body
  A contact force on the same body — equal here only because the book is in equilibrium.
• B. The normal contact force on the table from the floor — Unrelated Force
  Part of the table-floor interaction, unrelated to the book's weight.
• C. The book pushing down on the table — Wrong Interaction
  A real third-law pair, but with the table's push on the book, not with the Earth's pull.
• D. The weight of the table — Unrelated Force
  A different force on a different object entirely.

Common Mistake (⚠️):
Choosing the normal contact force from the table. Two forces on the same body that happen to balance are Newton's first law at work, not his third.

Takeaway (📌):
Third-law pairs: same interaction, same force type, opposite directions, different bodies. If both forces act on one object, it is not a pair.

Correct Answer: E. $40$

Fastest Approach (🚀):
$\dfrac{12}{240} = \dfrac{1}{20}$, so $N_s = \dfrac{800}{20} = 40$.

Matches Option E.

Step-by-Step Breakdown:

1. Write the transformer relationship

$\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$

2. Substitute and solve

$\dfrac{12}{240} = \dfrac{N_s}{800}$
$N_s = 800\times\dfrac{12}{240} = 800\times\dfrac{1}{20} = 40$

3. Sanity-check the direction first

The voltage is stepped down by a factor of $20$, so the secondary must have $20$ times fewer turns. Any answer larger than $800$ is wrong before the arithmetic is checked, which rules out $16\,000$ immediately.

4. How a transformer works at all

An alternating current in the primary produces a changing magnetic field in the core, which induces a voltage in the secondary. It is the changing field that matters — a transformer does nothing at all with direct current, because a steady field induces no voltage.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $16\,000$ — Ratio Inverted
  Multiplying by the ratio instead of dividing — a step-up answer.
• B. $20$ — Wrong Quantity
  Giving the turns ratio $20$ rather than the number of turns.
• C. $400$ — Arithmetic Error
  Dividing by $2$, or using $\dfrac{120}{240}$.
• D. $80$ — Arithmetic Error
  Dividing by $10$.

Common Mistake (⚠️):
Inverting the ratio and multiplying by $20$ instead of dividing, giving $16\,000$ turns. Checking the direction of the step first makes that impossible.

Takeaway (📌):
$\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$. Decide from the words whether the answer should be bigger or smaller before you compute it.

Correct Answer: A. mass $60\ \text{kg}$, weight $96\ \text{N}$

Fastest Approach (🚀):
Mass is unchanged: $60\ \text{kg}$.
$W = 60\times 1.6 = 96\ \text{N}$.

Matches Option A.

Step-by-Step Breakdown:

1. Mass does not change

Mass measures the amount of matter, in kilograms. Moving the astronaut to the Moon does not remove any matter, so the mass is still $60\ \text{kg}$.

2. Weight is a force, and it does change

Weight is the gravitational force on the mass:
$W = mg = 60\times 1.6 = 96\ \text{N}$

3. Compare with Earth

On Earth, $g = 10\ \text{N/kg}$ gives $W = 600\ \text{N}$. The Moon's weight is about a sixth of that, which is why the Apollo astronauts could bound around in heavy suits.

4. What that does and does not change

Jumping is easier because the weight is smaller. But bringing the astronaut to a stop after a push is exactly as hard as on Earth, because that depends on the mass, which is unchanged. Inertia travels; weight does not.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. mass $96\ \text{kg}$, weight $60\ \text{N}$ — Quantities Swapped
  Mass and weight swapped, and their units with them.
• C. mass $9.6\ \text{kg}$, weight $96\ \text{N}$ — Mass Changed
  Scaling the mass down as well as the weight.
• D. mass $60\ \text{kg}$, weight $600\ \text{N}$ — Wrong Field Strength
  Using Earth's $g$ for the weight.
• E. mass $10\ \text{kg}$, weight $16\ \text{N}$ — Mass Changed
  Dividing the mass by $6$ and then by $g$.

Common Mistake (⚠️):
Reducing the mass along with the weight. Kilograms measure matter and newtons measure force — a bathroom scale reads a force but is labelled in kilograms, which is where the confusion starts.

Takeaway (📌):
Mass in kilograms is the same everywhere. Weight in newtons is $mg$ and changes with location.

Correct Answer: C. $10\ \text{A}$

Fastest Approach (🚀):
$P = 240\times 0.5 = 120\ \text{W}$.
$I_s = \dfrac{120}{12} = 10\ \text{A}$.

Matches Option C.

Step-by-Step Breakdown:

1. Use the meaning of 'ideal'

An ideal transformer wastes no energy, so
$V_pI_p = V_sI_s$

2. Find the power

$P = 240\times 0.5 = 120\ \text{W}$

3. Divide by the secondary voltage

$I_s = \dfrac{120}{12} = 10\ \text{A}$

4. Read the trade-off

The voltage fell by a factor of $20$ and the current rose by a factor of $20$: $0.5\times 20 = 10\ \text{A}$. ✓ That is the entire purpose of a transformer — it trades one for the other while keeping the product fixed.

This is also why the national grid transmits at high voltage. Low current means little $I^{2}R$ heating in the cables, and transformers at each end convert back and forth.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $0.025\ \text{A}$ — Direction Reversed
  Scaling the current down with the voltage, so power vanishes.
• B. $0.5\ \text{A}$ — Ratio Ignored
  Assuming the current is unchanged.
• D. $20\ \text{A}$ — Ratio Error
  Using a turns ratio of $40$, or doubling the correct answer.
• E. $120\ \text{A}$ — Wrong Quantity
  Giving the power $120$ as a current.

Common Mistake (⚠️):
Assuming a step-down transformer reduces the current too, giving $0.025\ \text{A}$. Voltage and current move in opposite directions; if both fell, energy would vanish.

Takeaway (📌):
Ideal transformer: $V_pI_p = V_sI_s$. Whatever factor the voltage is divided by, the current is multiplied by.

Correct Answer: D. $10\ \text{cm}$

Fastest Approach (🚀):
Force $\times 2.5$, so extension $\times 2.5$.
$4\times 2.5 = 10\ \text{cm}$.

Matches Option D.

Step-by-Step Breakdown:

1. State the law

$F = kx$ — extension is directly proportional to the force, provided the elastic limit is not passed.

2. The ratio route

The force rises from $10\ \text{N}$ to $25\ \text{N}$, a factor of $2.5$. Proportionality means the extension rises by the same factor:
$4\times 2.5 = 10\ \text{cm}$

3. The spring constant route

$k = \dfrac{F}{x} = \dfrac{10}{0.04} = 250\ \text{N/m}$
$x = \dfrac{F}{k} = \dfrac{25}{250} = 0.1\ \text{m} = 10\ \text{cm}$

Both give $10\ \text{cm}$; the ratio route is faster, the constant route is better if the question continues.

4. Why the elastic limit is stated

Beyond it the graph stops being a straight line, the spring no longer returns to its original length, and proportionality fails entirely. Without that assurance the question could not be answered.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $8\ \text{cm}$ — Ratio Error
  Doubling rather than multiplying by 2.5.
• B. $100\ \text{cm}$ — Unit Error
  Using $k = 250$ but leaving the answer in the wrong unit.
• C. $6.25\ \text{cm}$ — Operation Error
  Dividing $25$ by $4$.
• E. $2.5\ \text{cm}$ — Ratio Inverted
  Dividing by the factor instead of multiplying.

Common Mistake (⚠️):
Working in centimetres while quoting $k$ in $\text{N/m}$, or doubling the extension because the force 'roughly doubled'. The factor is $2.5$, not $2$.

Takeaway (📌):
$F = kx$ is a direct proportion below the elastic limit, so scale the extension by the same factor as the force.

Correct Answer: A. $0.24\ \text{N}$

Fastest Approach (🚀):
$0.4\times 3 = 1.2$, then $1.2\times 0.2 = 0.24\ \text{N}$.

Matches Option A.

Step-by-Step Breakdown:

1. Apply the formula

$F = BIL = 0.4\times 3\times 0.2 = 0.24\ \text{N}$

2. Why the right angle matters

$F = BIL$ gives the maximum force, and applies only when the current is perpendicular to the field. Align the wire along the field instead and the force is zero — which is what Option E is testing.

3. Which way does it act

The force is perpendicular to both the current and the field, given by the left-hand rule: first finger field, second finger current, thumb motion. It is never along the wire.

4. What changes the force

Only three things: the field strength, the current, and the length of wire inside the field. Doubling any one of them doubles the force — which is how a motor is made more powerful.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $2.4\ \text{N}$ — Decimal Slip
  A factor of ten slip in the decimals.
• C. $1.2\ \text{N}$ — Factor Omitted
  Computing $BI$ and omitting the length.
• D. $0.6\ \text{N}$ — Factor Omitted
  Computing $3\times 0.2$ and omitting the field.
• E. $0\ \text{N}$ — Geometry Misread
  The force if the wire were parallel to the field, which it is not.

Common Mistake (⚠️):
Slipping a power of ten, giving $2.4\ \text{N}$. With three decimal factors it is worth grouping them as $(0.4\times 0.2)\times 3 = 0.08\times 3$ to keep the decimal place visible.

Takeaway (📌):
$F = BIL$ when field and current are perpendicular, zero when they are parallel, and the force is perpendicular to both.

Correct Answer: D. $2\ \text{m/s}$

Fastest Approach (🚀):
Momentum before $= 12\ \text{kg m/s}$.
$v = \dfrac{12}{6} = 2\ \text{m/s}$.

Matches Option D.

Step-by-Step Breakdown:

1. Momentum before

$p = m_1u_1+m_2u_2 = (2)(6)+(4)(0) = 12\ \text{kg m/s}$

The stationary trolley contributes nothing.

2. Momentum after

They move off together, so the combined mass is $2+4 = 6\ \text{kg}$:
$p = 6v$

3. Equate and solve

$6v = 12 \implies v = 2\ \text{m/s}$

4. Check the energy

Before: $\tfrac12(2)(6^{2}) = 36\ \text{J}$.
After: $\tfrac12(6)(2^{2}) = 12\ \text{J}$.

Two-thirds of the kinetic energy has gone — into deformation, sound and heat. This is an inelastic collision, and whenever bodies stick together the collision is inelastic. Momentum is still conserved exactly; kinetic energy is not, and assuming it is produces a different, wrong answer.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $3\ \text{m/s}$ — Wrong Mass
  Dividing by $4$ instead of by the combined mass of $6$.
• B. $1.5\ \text{m/s}$ — Wrong Mass
  Dividing $6$ by $4$, or using the wrong mass twice.
• C. $6\ \text{m/s}$ — Collision Ignored
  Assuming the moving trolley is unaffected.
• E. $12\ \text{m/s}$ — Wrong Quantity
  Giving the momentum $12\ \text{kg m/s}$ as a velocity.

Common Mistake (⚠️):
Conserving kinetic energy instead of momentum, or dividing by the moving mass alone rather than the combined mass.

Takeaway (📌):
Momentum is conserved in every collision. Kinetic energy is conserved only in elastic ones, and never when the bodies join.

Correct Answer: D. Holding the magnet stationary inside the coil

Fastest Approach (🚀):
Induction needs relative motion.
Stationary $\Rightarrow$ no voltage at all.

Matches Option D.

Step-by-Step Breakdown:

1. State the condition for induction

A voltage is induced only when a conductor cuts magnetic field lines, or equivalently when the field through a coil is changing. No change, no voltage.

2. Test each option

Faster motion: more field lines cut per second — larger voltage.
More turns: each turn contributes, so the voltages add — larger.
Stronger magnet: denser field lines, more cut per second — larger.
Moving the coil instead of the magnet: only the relative motion matters, so this works identically — larger.
Holding it stationary: nothing is changing, so the induced voltage falls to zero.

3. The result

Option D does not increase the voltage; it removes it. Note that the magnet being inside the coil is irrelevant — it is the motion, not the position, that induces anything.

4. Which way does the current flow

The induced current always opposes the change that caused it, so pushing a north pole in produces a north pole facing it, which pushes back. That opposition is why generating electricity takes work, and where the energy actually comes from.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. Pushing the magnet in more quickly — Does Increase
  More field lines cut per second, so a larger voltage.
• B. Using a coil with more turns — Does Increase
  Each turn adds its own induced voltage.
• C. Using a stronger magnet — Does Increase
  A denser field means more lines cut per second.
• E. Moving the coil towards the stationary magnet more quickly — Does Increase
  Only relative motion matters, so this is equivalent to moving the magnet.

Common Mistake (⚠️):
Assuming a magnet sitting inside a coil maintains a voltage. A steady field induces nothing, which is also why transformers do not work on direct current.

Takeaway (📌):
Induced voltage depends on the rate of change of field: speed, field strength and number of turns all raise it, and stopping the motion removes it.

Correct Answer: A. $10\ \text{m/s}$

Fastest Approach (🚀):
Impulse $= 15\times 4 = 60\ \text{kg m/s}$.
$v = \dfrac{60}{6} = 10\ \text{m/s}$.

Matches Option A.

Step-by-Step Breakdown:

1. Impulse equals change in momentum

$Ft = \Delta p = mv-mu$

2. Substitute

The object starts at rest, so $u = 0$:
$15\times 4 = 6v$
$60 = 6v$
$v = 10\ \text{m/s}$

3. The equivalent route

$a = \dfrac{F}{m} = \dfrac{15}{6} = 2.5\ \text{m/s}^{2}$, then $v = u+at = 0+2.5\times 4 = 10\ \text{m/s}$. ✓

Same answer. Note that $2.5$ appears among the options — it is the acceleration, and it is what you get by stopping one step early.

4. Why impulse is the useful form

In a real impact the force is not constant, and $Ft$ still works using the average force. That is why a longer collision time means a smaller force for the same change in momentum — the physics behind crumple zones, airbags and bending your knees when you land.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $2.5\ \text{m/s}$ — Wrong Quantity
  Giving the acceleration in $\text{m/s}^{2}$.
• C. $60\ \text{m/s}$ — Wrong Quantity
  Giving the impulse $60\ \text{kg m/s}$ as a speed.
• D. $3.6\ \text{m/s}$ — Formula Misuse
  Computing $\dfrac{6\times 15}{25}$ or another mis-formed ratio.
• E. $40\ \text{m/s}$ — Substitution Error
  Using $\dfrac{Ft}{m}$ with the mass and force interchanged.

Common Mistake (⚠️):
Quoting the acceleration $2.5\ \text{m/s}^{2}$ as a speed. Check the units of what you have actually computed before choosing.

Takeaway (📌):
$Ft = \Delta p$. For the same change in momentum, a longer impact time means a smaller force.

Correct Answer: A. $70\%$

Fastest Approach (🚀):
$\dfrac{350}{500} = 0.7 = 70\%$.

Matches Option A.

Step-by-Step Breakdown:

1. The definition

$\text{efficiency} = \dfrac{\text{useful energy transferred}}{\text{total energy supplied}}\times 100\%$

2. Substitute

$\dfrac{350}{500}\times 100 = 70\%$

3. Where the rest goes

$500-350 = 150\ \text{J}$ is dissipated — mostly as heat in the windings, plus some sound and friction. It is not destroyed: energy is conserved, and 'wasted' means transferred to a less useful store.

4. The check that costs nothing

Efficiency can never exceed $100\%$; that would create energy from nothing. Option B is $\dfrac{500}{350}$, the fraction the wrong way up, and it can be discarded on sight.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $143\%$ — Fraction Inverted
  Total over useful — impossible, as it exceeds 100%.
• C. $30\%$ — Wrong Quantity
  The percentage wasted rather than the efficiency.
• D. $150\%$ — Wrong Quantity
  Giving the $150\ \text{J}$ wasted as a percentage.
• E. $35\%$ — Arithmetic Error
  Dividing $350$ by $1000$.

Common Mistake (⚠️):
Inverting the fraction to get $143\%$. Any efficiency above $100\%$ is impossible and should be rejected before it is even computed.

Takeaway (📌):
Useful over total, never the reverse. The difference between the two is dissipated, usually as heat, and never destroyed.

Correct Answer: D. $20\ \text{m}$

Fastest Approach (🚀):
$h = \dfrac{v^{2}}{2g} = \dfrac{400}{20} = 20\ \text{m}$.

Matches Option D.

Step-by-Step Breakdown:

1. Energy at the start and the top

At launch the ball has kinetic energy $\tfrac12 mv^{2}$ and no height. At the top it is momentarily at rest, so all of it has become gravitational potential energy $mgh$.

2. Equate and cancel

$\tfrac12 mv^{2} = mgh$

The mass appears on both sides and cancels:
$h = \dfrac{v^{2}}{2g} = \dfrac{20^{2}}{2\times 10} = \dfrac{400}{20} = 20\ \text{m}$

3. Why no mass was given

The question omits the mass deliberately. Its absence is the signal that the energy equation will cancel it — a heavy ball and a light ball thrown at the same speed reach the same height.

4. The suvat cross-check

$v^{2} = u^{2}+2as$ with $v = 0$, $u = 20$, $a = -10$:
$0 = 400-20s \implies s = 20\ \text{m}$ ✓

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $40\ \text{m}$ — Factor Omitted
  Omitting the factor of $2$ in $2g$.
• B. $10\ \text{m}$ — Formula Misuse
  Using $\dfrac{v}{2}$ rather than $\dfrac{v^{2}}{2g}$.
• C. $2\ \text{m}$ — Formula Misuse
  Computing $\dfrac{20}{10}$.
• E. $200\ \text{m}$ — Factor Omitted
  Computing $\dfrac{v^{2}}{2}$ and ignoring $g$.

Common Mistake (⚠️):
Using $h = \dfrac{v^{2}}{g} = 40\ \text{m}$ and losing the factor of $2$, or dividing $v$ rather than $v^{2}$.

Takeaway (📌):
$\tfrac12 mv^{2} = mgh$ gives $h = \dfrac{v^{2}}{2g}$. When the mass is not given, it is because it cancels.

Correct Answer: B. $250\ \text{W}$

Fastest Approach (🚀):
$W = 200\times 15 = 3000\ \text{J}$.
$P = \dfrac{3000}{12} = 250\ \text{W}$.

Matches Option B.

Step-by-Step Breakdown:

1. Work done

$W = Fd = 200\times 15 = 3000\ \text{J}$

The force is along the direction of motion, so the full force counts. Had it been at an angle, only the component along the motion would do work.

2. Power

$P = \dfrac{W}{t} = \dfrac{3000}{12} = 250\ \text{W}$

3. The alternative one-liner

$P = Fv$, where $v = \dfrac{15}{12} = 1.25\ \text{m/s}$:
$P = 200\times 1.25 = 250\ \text{W}$ ✓

That form is worth knowing — it turns any 'force at constant speed' question into a single multiplication.

4. Units

Joules for work, watts for power, and one watt is one joule per second. Option A is the work, in joules, and is the obvious wrong pick for anyone who stops one step early.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $3000\ \text{W}$ — Incomplete
  The work done in joules, not the power.
• C. $160\ \text{W}$ — Substitution Error
  Computing $\dfrac{200\times 12}{15}$.
• D. $2400\ \text{W}$ — Wrong Distance
  Computing $200\times 12$.
• E. $0.8\ \text{W}$ — Inverted
  Dividing the speed by the force.

Common Mistake (⚠️):
Answering $3000$ — the work done rather than the power. The question asks for power, so the time must be used.

Takeaway (📌):
$W = Fd$ then $P = \dfrac{W}{t}$, or $P = Fv$ in one line when the speed is constant.