ESAT Physics Mock Module 3 Worked Solutions

Correct Answer: C. Horizontal circular field lines centred on the wire, running anticlockwise when viewed from above

Fastest Approach (🚀):
Right thumb up $\Rightarrow$ fingers curl anticlockwise as seen from above.
Field lines are closed horizontal circles centred on the wire.

Matches Option C.

Step-by-Step Breakdown:

1. What a magnetic field around a current looks like

A current-carrying wire produces field lines that close on themselves. They are circles centred on the wire, lying in planes perpendicular to it, and their strength falls off with distance from the wire. Nothing about a magnetic field ever starts or stops in space - there are no magnetic monopoles for the lines to begin or end on.

2. Fixing the direction with the grip rule

Point the right thumb along the conventional current, here vertically upwards, and let the fingers curl. The fingers now trace the field. Looking down on the wire from above, that curl is anticlockwise.

3. Checking the plane

The wire is vertical, so the perpendicular planes are horizontal. The circles therefore lie flat, one above another, all centred on the wire.

4. Why the sense matters

Reversing the current reverses the field, which is what makes the electromagnet in a relay or a scrapyard crane switchable. Getting the sense right is the whole content of this question.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. Straight field lines pointing radially outwards from the wire in all directions — Electric Field
  Radial lines are the shape of an electric field around a point charge. A magnetic field has no source to point away from.
• B. Circular field lines lying in a vertical plane that contains the wire — Wrong Plane
  The circles are perpendicular to the wire, not in a plane containing it. A plane containing a vertical wire is vertical; these circles are horizontal.
• D. Horizontal circular field lines centred on the wire, running clockwise when viewed from above — Sense Reversed
  Right shape, wrong sense - this is the field of a downward current. Curl the right hand with the thumb up and check which way the fingers go.
• E. A uniform field everywhere parallel to the wire and pointing upwards — Uniform Field
  A uniform field parallel to the current would exert no force on the current, and would have no source. Field lines circle a current; they never run along it.

Common Mistake (⚠️):
Reaching for the left hand. The left hand belongs to the motor effect (Fleming's left-hand rule, for the force on a current in a field); the field produced by a current is a right-hand rule.

Takeaway (📌):
The field of a straight current is closed circles round the wire, in planes perpendicular to it, with the sense given by the right-hand grip rule.

Correct Answer: B. $180\ \text{m}$

Fastest Approach (🚀):
Average velocity $= \dfrac{0+30}{2} = 15\ \text{m/s}$.
$s = 15 \times 12 = 180\ \text{m}$.

Matches Option B.

Step-by-Step Breakdown:

1. Why the average velocity is the quick route

When acceleration is uniform, velocity rises in a straight line, so the average velocity over the interval is simply the mean of the start and end values. That is what $s = \dfrac{u+v}{2}\,t$ says, and it avoids finding $a$ at all.

2. Substitute

$u = 0$, $v = 30\ \text{m/s}$, $t = 12\ \text{s}$:
$s = \dfrac{0+30}{2}\times 12 = 15 \times 12 = 180\ \text{m}$

3. The same answer from the graph

On a velocity-time graph this motion is a triangle of base $12\ \text{s}$ and height $30\ \text{m/s}$. Its area is $\tfrac12 \times 12 \times 30 = 180\ \text{m}$ - the same calculation wearing different clothes.

4. A bound worth carrying

The car can never cover more than $30 \times 12 = 360\ \text{m}$, which is what it would manage at the top speed the whole way, nor less than zero. The answer must sit in that range, and for a straight-line rise it lands exactly half way.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $90\ \text{m}$ — Halved Twice
  Half the answer - the average velocity halved a second time, or $\tfrac12 \times 15 \times 12$.
• C. $150\ \text{m}$ — Not Consistent
  Needs an average velocity of $12.5\ \text{m/s}$, which no consistent reading of the numbers gives.
• D. $250\ \text{m}$ — Not Consistent
  Between the two plausible wrong routes and matching neither. The average velocity is exactly $15\ \text{m/s}$.
• E. $360\ \text{m}$ — Final Speed Used
  $30 \times 12$ - the final velocity used throughout, as if the car never had to speed up.

Common Mistake (⚠️):
Using $s = vt$ with the final velocity, giving $360\ \text{m}$. That is the distance for a car already at $30\ \text{m/s}$ throughout, not one starting from rest.

Takeaway (📌):
For uniform acceleration, average velocity is the mean of the endpoints - and from rest that is just half the final speed.

Correct Answer: B. Air warmed near the floor expands, becomes less dense and rises, so a convection current carries heat through the whole room

Fastest Approach (🚀):
Warm air $\Rightarrow$ expands $\Rightarrow$ less dense $\Rightarrow$ rises.
Cool air sinks to replace it $\Rightarrow$ a current circulates the whole room.
A ceiling heater warms air that is already at the top, so nothing circulates.

Matches Option B.

Step-by-Step Breakdown:

1. What heating does to a fluid's density

Raising the temperature of a gas at constant pressure makes it expand. The mass is unchanged and the volume has grown, so from $\rho = \dfrac{m}{V}$ the density falls.

2. Why less dense fluid rises

A parcel of warm, low-density air is surrounded by cooler, denser air. The upthrust on it exceeds its weight, so it rises - the same buoyancy that lifts a hot-air balloon. Cooler air flows in underneath to take its place, is warmed in turn, and the cycle repeats.

3. Why the position matters so much

A heater at floor level sits at the bottom of that circulation, so every part of the room's air eventually passes over it. A heater at the ceiling warms air that is already as high as it can go: it stays there, and the cold air below is never drawn through the heater at all.

4. The same reasoning elsewhere

It is why a kettle's element sits at the bottom, why a fridge's freezer compartment is at the top, and why sea breezes reverse between day and night. One rule - warm fluid rises - explains all of them.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. Air conducts heat downwards much better than it conducts heat upwards — Conduction Invoked
  Conduction in a gas has no preferred direction, and air is a poor conductor in any case - which is why still air is used as an insulator.
• C. Thermal radiation travels upwards more readily than downwards, so a low heater radiates further — Radiation Directional
  Radiation travels equally well in every direction. It also does not require a medium, so the position of the heater in the room would make no difference to it.
• D. Cold air is less dense than warm air and sinks, pushing the warm air downwards — Density Reversed
  The density comparison is the wrong way round: cold air is denser than warm air, which is why it is the cold air that sinks.
• E. Convection cannot occur in gases, so only the position of the heater matters — Mechanism Denied
  Convection is exactly what does occur in gases - and in liquids. It is conduction that gases are poor at.

Common Mistake (⚠️):
Explaining it by conduction. Air is a poor conductor, which is precisely why still air in double glazing insulates; the bulk transfer of heat through a room is convection, not conduction.

Takeaway (📌):
Heat a fluid, it expands, its density falls and it rises. Put the heat source at the bottom so the circulation sweeps the whole volume.

Correct Answer: E. $15\ \text{Hz}$

Fastest Approach (🚀):
$f = \dfrac{v}{\lambda} = \dfrac{6}{0.4} = \dfrac{60}{4} = 15\ \text{Hz}$.

Matches Option E.

Step-by-Step Breakdown:

1. Rearrange the wave equation

$v = f\lambda \implies f = \dfrac{v}{\lambda}$

2. Substitute, clearing the decimal first

$f = \dfrac{6}{0.4} = \dfrac{60}{4} = 15\ \text{Hz}$

Multiplying top and bottom by $10$ turns an awkward decimal division into a whole-number one, which is worth doing every time in a non-calculator paper.

3. Check the direction of the change

The wavelength is less than a metre, so dividing by it must give a frequency larger than the speed's numerical value. Any answer at or below $6$ is wrong before the arithmetic is checked.

4. What the answer means physically

Fifteen complete waves pass a fixed point each second, and each carries $0.4\ \text{m}$ of rope past - which is the $6\ \text{m}$ per second the question started from.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $2.4\ \text{Hz}$ — Inverted Operation
  $6 \times 0.4$ - multiplied rather than divided. Check the units: $\text{m/s} \div \text{m}$ gives $\text{s}^{-1}$.
• B. $0.067\ \text{Hz}$ — Fraction Inverted
  $\dfrac{0.4}{6}$ - the fraction the wrong way up. This is the period in seconds, not the frequency.
• C. $24\ \text{Hz}$ — Decimal Slip
  Ten times the answer: the decimal cleared on the top only, giving $\dfrac{60}{2.5}$-style arithmetic.
• D. $6\ \text{Hz}$ — Wavelength Ignored
  The speed reported unchanged, as if the wavelength were $1\ \text{m}$.

Common Mistake (⚠️):
Multiplying instead of dividing, giving $2.4\ \text{Hz}$. The units settle it: $\text{m/s} \div \text{m}$ leaves $\text{s}^{-1}$, which is the hertz.

Takeaway (📌):
$f = v/\lambda$. Clear the decimal by scaling top and bottom before dividing.

Correct Answer: D. $25\ \text{kPa}$

Fastest Approach (🚀):
$p = \rho g h = 1000 \times 10 \times 2.5 = 25\,000\ \text{Pa}$.
$= 25\ \text{kPa}$.

Matches Option D.

Step-by-Step Breakdown:

1. Where $p = \rho g h$ comes from

Take a column of liquid of base area $A$ and height $h$. Its weight is $\rho A h g$, and pressure is force per unit area, so
$p = \dfrac{\rho A h g}{A} = \rho g h$

The area cancels. That is why the pressure at the bottom of a narrow pipe and a wide tank filled to the same depth is identical, and why the shape of the vessel is irrelevant.

2. Substitute

$p = 1000 \times 10 \times 2.5 = 25\,000\ \text{Pa}$

3. Convert to the prefixed unit

$25\,000\ \text{Pa} = 25\ \text{kPa}$, since $1\ \text{kPa} = 1000\ \text{Pa}$.

4. Sense check against atmospheric pressure

Atmospheric pressure is about $100\ \text{kPa}$. Two and a half metres of water giving a quarter of that is right: it takes roughly $10\ \text{m}$ of water to match the atmosphere, which is exactly why a suction pump cannot lift water higher than about ten metres.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $250\ \text{Pa}$ — Decimal Slip
  A hundred times too small - a decimal slip, most often the density read as $10\ \text{kg/m}^3$.
• B. $2.5\ \text{kPa}$ — Decimal Slip
  Ten times too small: $\rho g$ alone is $10\,000$, so multiplying by $2.5\ \text{m}$ cannot land at $2500$.
• C. $4\ \text{kPa}$ — Not Consistent
  No route through $\rho g h$ produces $4\ \text{kPa}$; the three given numbers multiply to $25\,000$ exactly.
• E. $250\ \text{kPa}$ — Decimal Slip
  Ten times too large. Note this exceeds two atmospheres, which $2.5\ \text{m}$ of water plainly cannot manage.

Common Mistake (⚠️):
Looking for the area of the tank base. It cancels out of $p = \rho g h$, so a question that gives no area is not missing anything.

Takeaway (📌):
Liquid pressure depends only on density, $g$ and depth. Ten metres of water is roughly one atmosphere - a useful anchor.

Correct Answer: D. $^{222}_{\ 86}\text{Rn}$

Fastest Approach (🚀):
$226 - 4 = 222$; $88 - 2 = 86$.
$Z = 86$ is radon.

Matches Option D.

Step-by-Step Breakdown:

1. What an alpha particle carries away

An alpha particle is a helium nucleus, $^{4}_{2}\text{He}$: two protons and two neutrons. It therefore removes $4$ from the mass number and $2$ from the atomic number.

2. Apply both conservation rules

Mass number: $226 - 4 = 222$

Atomic number: $88 - 2 = 86$

3. Name the element from $Z$, not the other way round

$Z = 86$ is radon. The atomic number is the element, so the symbol is read off after the subtraction rather than guessed alongside it.

4. The decay equation in full

$^{226}_{\ 88}\text{Ra} \rightarrow\ ^{222}_{\ 86}\text{Rn} + ^{4}_{2}\text{He}$

Both columns balance: $226 = 222 + 4$ and $88 = 86 + 2$. This is the real decay that fills basements with radon gas.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $^{222}_{\ 88}\text{Ra}$ — Charge Unchanged
  Mass number reduced but the atomic number left at $88$. Removing two protons must change the element.
• B. $^{226}_{\ 89}\text{Ac}$ — Beta Decay
  Atomic number raised by one and the mass unchanged - that is beta-minus decay, not alpha.
• C. $^{222}_{\ 87}\text{Fr}$ — One Proton Removed
  Only one proton removed. An alpha particle takes two, so $Z$ falls to $86$.
• E. $^{226}_{\ 86}\text{Rn}$ — Mass Unchanged
  Right element, but the mass number left at $226$. The four nucleons leave with the alpha particle.

Common Mistake (⚠️):
Changing one number and not the other. Alpha decay moves both; only gamma emission leaves both unchanged.

Takeaway (📌):
Alpha: $-4$ mass, $-2$ charge. Beta-minus: mass unchanged, $+1$ charge. Gamma: neither changes.

Correct Answer: C. $3000\ \text{W}$

Fastest Approach (🚀):
$E = mgh = 400 \times 10 \times 15 = 60\,000\ \text{J}$.
$P = \dfrac{E}{t} = \dfrac{60\,000}{20} = 3000\ \text{W}$.

Matches Option C.

Step-by-Step Breakdown:

1. Energy given to the load

Lifting at a steady speed means no kinetic energy is gained, so all the useful work goes into gravitational potential energy:
$E = mgh = 400 \times 10 \times 15 = 60\,000\ \text{J}$

2. Power is that energy per second

$P = \dfrac{E}{t} = \dfrac{60\,000}{20} = 3000\ \text{W} = 3\ \text{kW}$

3. The same answer as force times speed

The load rises at $\dfrac{15}{20} = 0.75\ \text{m/s}$ against a weight of $400 \times 10 = 4000\ \text{N}$, and $P = Fv = 4000 \times 0.75 = 3000\ \text{W}$. Two routes, one answer - worth knowing both, because $P = Fv$ is faster whenever a steady speed is given directly.

4. Why 'useful'

The crane's motor draws more than $3\ \text{kW}$: some goes to friction, heat and moving the crane's own arm. The question asks only for the power delivered to the load, which is the $mgh$ figure.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $300\ \text{W}$ — Decimal Slip
  A tenth of the answer, from $g$ used as $1$ or the height read as $1.5\ \text{m}$.
• B. $1200\ \text{W}$ — Not Consistent
  $400 \times 15 \div 5$-style arithmetic; no consistent route reaches it. Check with $P = Fv = 4000 \times 0.75$.
• D. $6000\ \text{W}$ — Time Halved
  Twice the answer - the time taken as $10\ \text{s}$ rather than $20\ \text{s}$.
• E. $60\,000\ \text{W}$ — Energy Reported
  The energy in joules, not the power. Divide by the $20\ \text{s}$ to get watts.

Common Mistake (⚠️):
Reporting the energy rather than the power. $60\,000$ is joules, not watts - the division by $20\ \text{s}$ is the step that turns one into the other.

Takeaway (📌):
Power is energy per second. Compute the energy in full, then divide by the time exactly once.

Correct Answer: D. $1.5\ \text{A}$

Fastest Approach (🚀):
Equal pair in parallel: $\dfrac{4}{2} = 2\,\Omega$.
Series total: $6 + 2 = 8\,\Omega$.
$I = \dfrac{12}{8} = 1.5\ \text{A}$.

Matches Option D.

Step-by-Step Breakdown:

1. Collapse the parallel pair

For two resistors in parallel, $\dfrac{1}{R_p} = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}$, so $R_p = 2\,\Omega$. For equal resistors there is a shortcut worth having: $n$ equal resistors $R$ in parallel give $\dfrac{R}{n}$, so a pair of $4\,\Omega$ gives $2\,\Omega$ immediately.

2. Add what is now in series

The $6\,\Omega$ resistor and the $2\,\Omega$ block carry the same current, so they add:
$R_{\text{total}} = 6 + 2 = 8\,\Omega$

3. Apply Ohm's law to the whole loop

$I = \dfrac{V}{R} = \dfrac{12}{8} = 1.5\ \text{A}$

4. Sanity-check the split

That $1.5\ \text{A}$ divides equally between the two $4\,\Omega$ branches, $0.75\ \text{A}$ each. The voltages then read $1.5 \times 6 = 9\ \text{V}$ across the series resistor and $1.5 \times 2 = 3\ \text{V}$ across the pair, which sum to the battery's $12\ \text{V}$ exactly.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $0.86\ \text{A}$ — Parallel Added
  $\dfrac{12}{14}$ - the parallel pair added as though it were in series. A parallel combination is always smaller than either branch.
• B. $1.0\ \text{A}$ — Wrong Total
  Needs $12\,\Omega$ in total. Neither $6+4+2$ nor any other reading of the network gives that.
• C. $1.2\ \text{A}$ — Branch Ignored
  Needs $10\,\Omega$, which is $6+4$: one of the two parallel resistors counted and the other ignored.
• E. $2.0\ \text{A}$ — Block Ignored
  $\dfrac{12}{6}$ - the parallel block left out altogether, as if the series resistor were the whole circuit.

Common Mistake (⚠️):
Adding the parallel pair as $4 + 4 = 8$, giving $14\,\Omega$ and $0.86\ \text{A}$. Parallel resistance is always smaller than the smallest branch - if your combined value went up, the rule was applied the wrong way round.

Takeaway (📌):
Parallel first, series second, Ohm's law last. Equal resistors in parallel divide by how many there are.

Correct Answer: A. The gravitational pull of the book on the Earth

Fastest Approach (🚀):
Weight $=$ Earth pulls book (gravity).
Partner $=$ book pulls Earth (gravity), equal and opposite.

Matches Option A.

Step-by-Step Breakdown:

1. Write the force as 'X pulls Y'

Every force is one body acting on another. The book's weight is the Earth pulling the book. Naming it that way makes the partner almost automatic: swap the two bodies and keep the type of force.

2. Swap the bodies

Partner $=$ the book pulling the Earth, gravitationally, with exactly the same magnitude and the opposite direction. The Earth barely responds because its mass is enormous, but the force on it is the same size.

3. Why the contact force is not the partner

The normal force of the table on the book is also equal and opposite to the weight here - but only because the book happens to be in equilibrium. Tilt the table, or put the book in a lift that accelerates, and they stop being equal, while the true third-law pair stays equal always. They are also the wrong kind of force (contact, not gravitational) and they act on the same body.

4. The test to apply

A third-law pair is: same type of force, two different bodies, roles exchanged, and equal in size under all circumstances. Two forces on the same body are never a third-law pair.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. The normal contact force of the table on the book — Same Body
  Equal and opposite here, but it acts on the same body and is a contact force rather than a gravitational one. In an accelerating lift it would no longer match the weight; a third-law partner always does.
• C. The normal contact force of the book on the table — Wrong Pair
  This is a genuine third-law partner - of the table's normal force on the book, not of the book's weight. Right law, wrong pair.
• D. The weight of the table — Unrelated Force
  A force on a different object entirely, and unrelated in size to the book's weight.
• E. The friction between the book and the table — Wrong Force Type
  Zero here, since nothing pushes the book sideways, and a contact force rather than a gravitational one.

Common Mistake (⚠️):
Choosing the normal contact force because it is equal and opposite. Balanced forces on one body come from the first law; the third law always involves two bodies.

Takeaway (📌):
Say the force as 'A acts on B', then swap A and B keeping the force type. If both forces act on the same object, it is not a third-law pair.

Correct Answer: A. $420\ \text{kJ}$

Fastest Approach (🚀):
$\Delta\theta = 70 - 20 = 50\ ^\circ\text{C}$.
$E = 2.0 \times 4200 \times 50 = 420\,000\ \text{J} = 420\ \text{kJ}$.

Matches Option A.

Step-by-Step Breakdown:

1. Identify the temperature change

$\Delta\theta = 70 - 20 = 50\ ^\circ\text{C}$. The formula uses the change, never the final reading - which is also why a change in $^\circ$C and a change in kelvin are numerically the same and either may be used.

2. Substitute

$E = mc\Delta\theta = 2.0 \times 4200 \times 50$

3. Multiply in a convenient order

$2.0 \times 4200 = 8400$, and $8400 \times 50 = 420\,000\ \text{J}$.

Doing the $\times 50$ last keeps every step a small multiplication, which matters in a paper with no calculator.

4. Convert

$420\,000\ \text{J} = 420\ \text{kJ}$. As a sanity check, a domestic $3\ \text{kW}$ kettle would take $\dfrac{420\,000}{3000} = 140\ \text{s}$ to deliver it, which is about right for two litres.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $588\ \text{kJ}$ — Change Not Used
  $2.0 \times 4200 \times 70$ - the final temperature used in place of the $50\ ^\circ\text{C}$ rise.
• C. $210\ \text{kJ}$ — Halved
  Half the answer: the mass taken as $1.0\ \text{kg}$, or the rise as $25\ ^\circ\text{C}$.
• D. $42\ \text{kJ}$ — Decimal Slip
  A factor of ten too small - a decimal slip in $4200 \times 50$.
• E. $4.2\ \text{kJ}$ — Decimal Slip
  A hundred times too small. $2.0 \times 4200$ alone is already $8400\ \text{J}$ per degree.

Common Mistake (⚠️):
Using $70\ ^\circ\text{C}$ instead of the $50\ ^\circ\text{C}$ rise, giving $588\ \text{kJ}$. Only a change in temperature has any meaning in $E = mc\Delta\theta$.

Takeaway (📌):
$E = mc\Delta\theta$, with $\Delta\theta$ the rise. Multiply the awkward number last.

Correct Answer: A. The frequency is unchanged and the wavelength decreases

Fastest Approach (🚀):
Crests cannot accumulate at a boundary $\Rightarrow$ $f$ unchanged.
$v$ falls and $f$ is fixed $\Rightarrow$ $\lambda = v/f$ falls.

Matches Option A.

Step-by-Step Breakdown:

1. Why the frequency cannot change

Frequency is set by the source. Whatever rate crests arrive at the boundary, they must leave it at the same rate: if they left more slowly, crests would build up at the line without limit, and if faster, crests would have to be created there. Neither happens, so $f$ is identical on both sides.

2. What that forces on the wavelength

From $v = f\lambda$, with $f$ fixed, $\lambda \propto v$. The wave slows, so the wavelength shortens in exact proportion. Halve the speed and the wavefronts sit half as far apart.

3. Why the wave also bends

If the wavefronts meet the boundary at an angle, the end that enters the shallow water first slows first, so the front pivots. That is refraction, and the change in wavelength is what causes it - not a separate effect.

4. The same result for light

Light entering glass slows and its wavelength shortens, while its frequency - and therefore its colour - is unchanged. A red laser is still red inside glass.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. Both the frequency and the wavelength decrease — Frequency Changed
  A falling frequency would mean crests disappearing at the boundary. The source is still producing them at the same rate.
• C. The frequency decreases and the wavelength is unchanged — Reversed
  Exactly the wrong way round: it is the frequency that is fixed and the wavelength that changes.
• D. Both are unchanged; only the direction of travel changes — Wavelength Fixed
  The direction does change, but so does the wavelength - and the change in wavelength is the reason the direction changes at all.
• E. The frequency increases and the wavelength decreases — Frequency Changed
  The wavelength is right, but nothing can raise the frequency; that would require crests to be created at the boundary.

Common Mistake (⚠️):
Assuming a slower wave must be a lower-frequency one. Speed and frequency are independent here: the medium sets the speed, the source sets the frequency, and the wavelength is whatever $v/f$ demands.

Takeaway (📌):
Across any boundary, frequency is conserved. Whatever the speed does, the wavelength does in proportion.

Correct Answer: C. $2700\ \text{kg/m}^3$

Fastest Approach (🚀):
$V = 100\ \text{cm}^3$.
$\rho = \dfrac{270}{100} = 2.7\ \text{g/cm}^3$.
$\times 1000 \Rightarrow 2700\ \text{kg/m}^3$.

Matches Option C.

Step-by-Step Breakdown:

1. Volume in the units given

$V = 5 \times 4 \times 5 = 100\ \text{cm}^3$

2. Density in those same units

$\rho = \dfrac{m}{V} = \dfrac{270}{100} = 2.7\ \text{g/cm}^3$

3. Convert once, at the end

$1\ \text{g/cm}^3 = 1000\ \text{kg/m}^3$, because a gram is $10^{-3}$ of a kilogram while a cubic centimetre is $10^{-6}$ of a cubic metre, and $\dfrac{10^{-3}}{10^{-6}} = 10^{3}$.

$\rho = 2.7 \times 1000 = 2700\ \text{kg/m}^3$

4. Recognise the material

$2700\ \text{kg/m}^3$ is aluminium. Water is $1000$, steel about $7800$ and lead about $11\,300$ - a small table worth carrying, because it catches a factor-of-ten slip instantly.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $27\ \text{kg/m}^3$ — Conversion Slip
  A hundred times too small - the $\times 1000$ conversion applied as $\times 10$, or the volume taken as $1000\ \text{cm}^3$.
• B. $270\ \text{kg/m}^3$ — Conversion Slip
  Ten times too small: this is $\dfrac{270}{100}$ with only a $\times 100$ conversion applied.
• D. $27\,000\ \text{kg/m}^3$ — Conversion Slip
  Ten times too large - the conversion applied twice over, or the volume read as $10\ \text{cm}^3$.
• E. $0.37\ \text{kg/m}^3$ — Fraction Inverted
  $\dfrac{100}{270}$ - mass and volume the wrong way round. Density is mass per unit volume.

Common Mistake (⚠️):
Converting the centimetres to metres one dimension at a time and losing a power of ten. Cubing a length conversion means cubing the factor: $10^{-2}$ becomes $10^{-6}$.

Takeaway (📌):
Compute in g/cm³, then multiply by $1000$ for kg/m³. Water is the anchor: $1\ \text{g/cm}^3 = 1000\ \text{kg/m}^3$.

Correct Answer: B. $8\ \text{days}$

Fastest Approach (🚀):
$800 \to 400 \to 200 \to 100$: three halvings.
$\dfrac{24}{3} = 8$ days.

Matches Option B.

Step-by-Step Breakdown:

1. Count the halvings rather than the ratio

The activity fell by a factor of $8$, and $8 = 2^3$. Writing the chain out makes it plain:

$800 \to 400 \to 200 \to 100$

Three steps, so three half-lives have elapsed.

2. Divide the elapsed time

$t_{1/2} = \dfrac{24}{3} = 8\ \text{days}$

3. Check it forwards

Starting at $800\ \text{Bq}$: after $8$ days $400$, after $16$ days $200$, after $24$ days $100$. The figures reproduce the question exactly.

4. Why the factor of $8$ is the thing to spot

Ratios that are powers of two - $2$, $4$, $8$, $16$, $32$ - turn every half-life question into a counting exercise with no logarithms. It is worth recognising them on sight.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $6\ \text{days}$ — Halvings Miscounted
  $\dfrac{24}{4}$ - four halvings counted, which would take the activity down to $50\ \text{Bq}$.
• C. $12\ \text{days}$ — Halvings Miscounted
  $\dfrac{24}{2}$ - two halvings, which would leave $200\ \text{Bq}$.
• D. $3\ \text{days}$ — Ratio Not Count
  $\dfrac{24}{8}$ - the elapsed time divided by the ratio rather than by the number of halvings.
• E. $24\ \text{days}$ — Whole Interval
  The whole elapsed time, which would be the half-life only if the activity had fallen to $400\ \text{Bq}$.

Common Mistake (⚠️):
Dividing $24$ by the ratio $8$ to get $3$ days. It is the number of halvings that divides the time, and eight is the ratio, not the count.

Takeaway (📌):
Write the halving chain out. The number of arrows is what divides the elapsed time.

Correct Answer: E. $6\ \text{V}$

Fastest Approach (🚀):
$\dfrac{N_s}{N_p} = \dfrac{50}{2000} = \dfrac{1}{40}$.
$V_s = \dfrac{240}{40} = 6\ \text{V}$.

Matches Option E.

Step-by-Step Breakdown:

1. Decide the direction before touching the numbers

The secondary has far fewer turns than the primary, so this is a step-down transformer and the secondary voltage must be well below $240\ \text{V}$. That single observation eliminates option A immediately.

2. Simplify the turns ratio

$\dfrac{N_s}{N_p} = \dfrac{50}{2000} = \dfrac{1}{40}$

3. Apply it to the voltage

$V_s = V_p \times \dfrac{N_s}{N_p} = \dfrac{240}{40} = 6\ \text{V}$

4. What the current does

For an ideal transformer the power is the same on both sides, so the secondary current is $40$ times the primary current. Stepping the voltage down steps the current up by the same factor - which is exactly why the transformer in a welding set can deliver hundreds of amps from a domestic supply.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $9600\ \text{V}$ — Ratio Inverted
  $240 \times 40$ - the turns ratio inverted, giving a step-up from a coil with fewer secondary turns.
• B. $60\ \text{V}$ — Ratio Misread
  A ratio of $\tfrac14$, which would need $500$ secondary turns rather than $50$.
• C. $24\ \text{V}$ — Ratio Misread
  A ratio of $\tfrac{1}{10}$: the $2000$ read as $500$, or a factor of four lost in simplifying.
• D. $12\ \text{V}$ — Ratio Misread
  A ratio of $\tfrac{1}{20}$ - half the correct step-down, from $\tfrac{50}{1000}$.

Common Mistake (⚠️):
Inverting the ratio and getting $9600\ \text{V}$. A coil with fewer turns can only produce a smaller voltage; check the direction before the arithmetic.

Takeaway (📌):
$\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p}$. Say 'step-up' or 'step-down' out loud first and the ratio cannot go in upside down.

Correct Answer: E. $1\ \text{m/s}$

Fastest Approach (🚀):
Before: $p = 0.5 \times 4 = 2\ \text{kg\,m/s}$.
After: total mass $= 2\ \text{kg}$.
$v = \dfrac{2}{2} = 1\ \text{m/s}$.

Matches Option E.

Step-by-Step Breakdown:

1. Momentum before the collision

Only the first trolley is moving:
$p = mv = 0.5 \times 4 = 2\ \text{kg\,m/s}$

The stationary trolley contributes nothing.

2. Momentum after

They stick, so they move as one body of mass $0.5 + 1.5 = 2\ \text{kg}$ at a single speed $v$:
$p = 2v$

3. Equate and solve

External forces are negligible during the impact, so momentum is conserved:
$2v = 2 \implies v = 1\ \text{m/s}$

4. What happened to the energy

Before: $\tfrac12(0.5)(4^2) = 4\ \text{J}$. After: $\tfrac12(2)(1^2) = 1\ \text{J}$. Three quarters of the kinetic energy has gone into deformation and heat. That is normal for a collision in which the bodies stick, and it is exactly why kinetic energy cannot be used to find $v$.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $4\ \text{m/s}$ — No Interaction
  The incoming trolley's speed, unchanged - which would need the second trolley to have no mass at all.
• B. $3\ \text{m/s}$ — Not Consistent
  Neither the momentum nor the energy calculation produces this; the combined mass is four times the moving one, so the speed must drop by a factor of four.
• C. $2\ \text{m/s}$ — Mass Understated
  Divides the initial momentum by $1\ \text{kg}$, or halves the incoming speed - the stationary trolley's $1.5\ \text{kg}$ under-counted.
• D. $1.5\ \text{m/s}$ — Wrong Mass
  Uses only the stationary trolley's mass in the final step. It is the combined $2\ \text{kg}$ that moves afterwards.

Common Mistake (⚠️):
Trying to conserve kinetic energy. In any collision where the objects stick together, kinetic energy is always lost - momentum is the only quantity that survives.

Takeaway (📌):
Stick-together collisions: total momentum before, divided by total mass after. Never energy.

Correct Answer: D. The metal conducts energy away from the hand far faster than the wood does, so the hand loses energy more quickly

Fastest Approach (🚀):
Same room, many hours $\Rightarrow$ same temperature.
Metal conducts far better $\Rightarrow$ faster energy loss from the hand $\Rightarrow$ feels colder.

Matches Option D.

Step-by-Step Breakdown:

1. They are at the same temperature

Both have been in the room long enough to reach thermal equilibrium with it. A thermometer touched to each would read the same. So temperature cannot be the explanation.

2. What 'feeling cold' actually measures

Nerve endings in the skin respond to the rate at which energy leaves the hand, not to the temperature of what is being touched. Anything that removes energy quickly feels cold.

3. Why metal removes it quickly

Metals conduct thermally through their free electrons, which move through the lattice and carry energy with them. Wood has no free electrons and conducts only through slow lattice vibrations, so it is a good insulator. The difference in conductivity is a factor of several hundred.

4. The check that settles it

In a hot room, above body temperature, the metal knob feels hotter than the wood - the same conduction now working in the other direction. A property that reverses with the direction of heat flow cannot be temperature.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. The metal is at a lower temperature than the wood — Temperature Assumed
  After hours in the same room both are at room temperature. A thermometer confirms it; the hand does not.
• B. The metal has a lower specific heat capacity, so it cools the room around it — Wrong Property
  An object cannot cool a room it is in equilibrium with, and specific heat capacity governs how much energy changes its temperature, not how fast energy flows through it.
• C. The metal emits more thermal radiation towards the hand — Radiation Invoked
  Radiation from a room-temperature object flows both ways and is small either way. Polished metal is in fact a poor emitter, so this would make it feel warmer, not colder.
• E. The wood is the better conductor, but its rough surface traps a layer of warm air — Reversed
  The conductivities are the wrong way round: metal is the good conductor and wood the insulator.

Common Mistake (⚠️):
Assuming the metal really is colder. Left in the same room for hours, everything reaches the same temperature; what differs is how fast each material moves energy.

Takeaway (📌):
Touch senses the rate of energy transfer, not temperature. Good conductors feel cold in a cool room and hot in a hot one.

Correct Answer: C. Ultraviolet, visible light, infrared

Fastest Approach (🚀):
Order by increasing $\lambda$: gamma, X-ray, UV, visible, infrared, microwave, radio.
UV $\to$ visible $\to$ infrared reads left to right along it.

Matches Option C.

Step-by-Step Breakdown:

1. Fix the spectrum in one direction

Written by increasing wavelength (and so decreasing frequency and photon energy):

gamma rays
X-rays
ultraviolet
visible light
infrared
microwaves

• radio waves

2. Read the option in that direction

Ultraviolet, visible light, infrared are three consecutive entries taken left to right, so their wavelengths increase.

3. Anchor it with a fact you can check

Ultraviolet causes sunburn and infrared is felt as warmth. Higher energy per photon means shorter wavelength, so ultraviolet must be the short end and infrared the long one, with visible light between - which is why the visible band is named as it is between the two.

4. Why the order is worth memorising once

Almost every spectrum question is this list read in one direction or the other. Learn it in one direction only and reverse it deliberately when asked, rather than trying to hold both.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. Infrared, visible light, ultraviolet — Direction Reversed
  The correct three regions in decreasing order of wavelength. Infrared is the longest of them, not the shortest.
• B. X-rays, gamma rays, microwaves — Not Monotonic
  Not monotonic: X-rays have a longer wavelength than gamma rays, so this list falls and then rises steeply.
• D. Radio waves, microwaves, infrared — Direction Reversed
  Decreasing: radio waves are the longest of all, and infrared the shortest of the three.
• E. Visible light, ultraviolet, X-rays — Direction Reversed
  Decreasing again - X-rays are far shorter than visible light.

Common Mistake (⚠️):
Answering with the right three regions in the wrong direction. Decide first whether the question wants wavelength or frequency, and increasing or decreasing - they invert one another.

Takeaway (📌):
One list, one direction: gamma to radio is increasing wavelength and decreasing energy. Reverse it consciously when the question does.

Correct Answer: A. The pressure doubles, because the molecules move faster and so strike the walls both harder and more often

Fastest Approach (🚀):
Rigid container, fixed mass $\Rightarrow p \propto T$.
$T$ doubled $\Rightarrow p$ doubled.
Mechanism: faster molecules, harder and more frequent impacts.

Matches Option A.

Step-by-Step Breakdown:

1. The relationship

For a fixed mass of ideal gas at constant volume, pressure is directly proportional to absolute temperature. Doubling $T$ in kelvin doubles $p$. The question says 'absolute' precisely so that this applies without any conversion.

2. Why, molecularly

Temperature measures the mean kinetic energy of the molecules. Raise it and:

each molecule arrives at the wall with more momentum, so each collision delivers a larger impulse; and
the molecules cross the container more quickly, so collisions happen more often.

Both effects push the pressure up, and together they give the proportionality.

3. Why the factor is two and not four

Mean kinetic energy is proportional to $T$, and kinetic energy goes as $v^2$, so the speed rises by $\sqrt2$ rather than by $2$. Impulse per collision rises by $\sqrt2$ and collision frequency by $\sqrt2$, and $\sqrt2 \times \sqrt2 = 2$. The two square roots multiplying to give the factor is exactly where option D goes wrong.

4. What has not changed

The number of molecules, the volume and the size of the molecules are all unchanged. Nothing about heating a gas makes its molecules bigger.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. The pressure halves, because the molecules spread further apart — Direction Reversed
  The container is rigid, so the molecules cannot spread out at all; and heating a gas raises its pressure rather than lowering it.
• C. The pressure is unchanged, because the volume has not changed — Change Denied
  Constant volume fixes the relationship between $p$ and $T$; it does not stop the pressure changing. That is how a pressure-gauge thermometer works.
• D. The pressure quadruples, because pressure depends on the square of the molecular speed — Square Applied
  Mean kinetic energy is proportional to $T$, so speed rises only by $\sqrt2$. Impulse and frequency each gain $\sqrt2$, giving $2$ overall.
• E. The pressure doubles, because the molecules themselves become larger — Wrong Mechanism
  Right factor, wrong mechanism. Molecules do not change size with temperature; they simply move faster.

Common Mistake (⚠️):
Using temperature in degrees Celsius. The proportionality holds only for absolute temperature - doubling $20\,^\circ\text{C}$ to $40\,^\circ\text{C}$ is a rise of just $293$ to $313\ \text{K}$, about 7%.

Takeaway (📌):
Constant volume: $p \propto T$ in kelvin. The mechanism is harder and more frequent collisions, each contributing a factor of $\sqrt2$.

Correct Answer: A. Vertically downwards

Fastest Approach (🚀):
Force is perpendicular to both current and field $\Rightarrow$ vertical.
Left hand: first finger east, second finger north $\Rightarrow$ thumb down.

Matches Option A.

Step-by-Step Breakdown:

1. Establish that there is a force at all

A current in a magnetic field experiences a force unless it runs parallel to the field. Here the current is north and the field east - perpendicular - so the force is at its maximum.

2. Narrow it to two possibilities

The force is always perpendicular to both the current and the field. Both of those are horizontal and at right angles to each other, so the only remaining direction is vertical: up or down.

3. Choose between them with the left hand

Fleming's left-hand rule, with the three fingers mutually perpendicular:

• First finger - Field: point it east.
• seCond finger - Current: point it north.
• thuMb - Motion (the force): it now points downwards.

4. Why the left hand here

The left hand is for the force on a current in a field - the motor effect. The right hand belongs to the field produced by a current, and to induced current in a generator. Using the wrong hand reverses the answer, which is why option B is on the list.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. Vertically upwards — Wrong Hand
  The right hand used in place of the left, which reverses the force. The motor effect is a left-hand rule.
• C. Horizontally, due west — Not Perpendicular
  Antiparallel to the field, which the force can never be - it is perpendicular to both current and field.
• D. Horizontally, due south — Not Perpendicular
  Antiparallel to the current, which the force can never be either.
• E. There is no force on the wire — Force Denied
  A current parallel to the field feels no force, but here they are at right angles, which is where the force is largest.

Common Mistake (⚠️):
Using the right hand and getting 'upwards'. Motor effect is left hand; the right hand is for the field around a current and for induction.

Takeaway (📌):
Force is perpendicular to both current and field, so with both horizontal the answer is vertical. Fleming's left hand picks which way.

Correct Answer: B. $25\ \text{N}$

Fastest Approach (🚀):
Extension $\times \dfrac{10}{4} = \times 2.5$.
Load $= 10 \times 2.5 = 25\ \text{N}$.

Matches Option B.

Step-by-Step Breakdown:

1. What Hooke's law asserts

Within the elastic limit, $F = kx$: the extension is directly proportional to the load. Directly proportional means the graph is a straight line through the origin, so ratios can be scaled without finding $k$ at all.

2. Scale the extension

$\dfrac{10\ \text{cm}}{4\ \text{cm}} = 2.5$

3. Scale the load by the same factor

$F = 10 \times 2.5 = 25\ \text{N}$

4. The same answer through $k$

$k = \dfrac{10\ \text{N}}{0.04\ \text{m}} = 250\ \text{N/m}$, and $F = 250 \times 0.10 = 25\ \text{N}$. Identical, but with two unit conversions that the proportion avoids entirely.

5. Why the elastic limit is stated

Beyond it the graph curves and the proportion fails, so a question that scales loads like this must say the spring stays elastic - as this one does.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $4\ \text{N}$ — Quantity Confused
  The extension in centimetres restated as a force. The load and the extension are different quantities.
• C. $40\ \text{N}$ — Wrong Operation
  $10 \times 4$ - the two given numbers multiplied, rather than the ratio applied.
• D. $100\ \text{N}$ — Wrong Ratio
  A factor of ten, as if the extension had gone from $1\ \text{cm}$ to $10\ \text{cm}$.
• E. $2.5\ \text{N}$ — Ratio Reported
  $\dfrac{10}{4}$ - the ratio reported as the answer, without applying it to the $10\ \text{N}$ load.

Common Mistake (⚠️):
Adding the difference instead of scaling: extension up by $6\ \text{cm}$ does not mean load up by $6\ \text{N}$. Proportional means multiply, not add.

Takeaway (📌):
Hooke's law is a straight line through the origin, so scale factors carry straight across. Only convert to metres if you actually want $k$.

Correct Answer: E. The black can cools faster, because a matt black surface is the better emitter of infrared radiation

Fastest Approach (🚀):
Matt black $\Rightarrow$ best absorber $\Rightarrow$ best emitter.
Best emitter $\Rightarrow$ loses energy fastest $\Rightarrow$ cools fastest.

Matches Option E.

Step-by-Step Breakdown:

1. Absorption and emission go together

A surface that absorbs infrared radiation well also emits it well - the two properties are governed by the same physics and cannot be separated. Matt black is the best at both; polished silver is poor at both.

2. Apply it to cooling

Cooling here is the water losing energy to its surroundings. The black can radiates more energy per second at any given temperature, so its temperature falls faster.

3. Where the intuition goes wrong

'Shiny reflects heat away' feels like it should mean faster cooling, but reflection describes what the surface does with radiation arriving from outside. What sets the cooling rate is what the surface emits, and a reflective surface emits very little. It is exactly why a vacuum flask is silvered on both faces.

4. The same experiment run the other way

Fill the cans with cold water and stand them in sunlight and the black one warms fastest - the same property, now working as absorption. That symmetry is the point of the question.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. The silver can cools faster, because a shiny surface reflects heat away — Reflection Invoked
  Reflection concerns radiation arriving from outside. A shiny surface is a poor emitter, so it cools more slowly - which is why flasks are silvered.
• B. Both cool at the same rate, because they hold the same water at the same temperature — Surface Ignored
  The contents are identical, but the rate of energy loss is set by the surface, and the surfaces differ.
• C. The silver can cools faster, because it is the better absorber — Reversed
  The absorption is the wrong way round: matt black absorbs best, polished silver worst.
• D. The black can cools more slowly, because black surfaces absorb radiation — Emission Ignored
  It absorbs best and emits best, so it cools fastest, not slowest.

Common Mistake (⚠️):
Treating 'reflects heat away' as a cooling mechanism. Reflection concerns incoming radiation; the rate a hot object cools is set by how well it emits.

Takeaway (📌):
Good absorber, good emitter; poor absorber, poor emitter. Matt black cools fastest and warms fastest.

Correct Answer: C. $1.0\ \text{s}$

Fastest Approach (🚀):
Round trip: $2 \times 170 = 340\ \text{m}$.
$t = \dfrac{340}{340} = 1.0\ \text{s}$.

Matches Option C.

Step-by-Step Breakdown:

1. The distance the sound actually covers

The clap travels to the wall, reflects, and returns to the listener. The path length is therefore twice the separation:
$s = 2 \times 170 = 340\ \text{m}$

2. Divide by the speed

$t = \dfrac{s}{v} = \dfrac{340}{340} = 1.0\ \text{s}$

3. Why the numbers are chosen as they are

$170\ \text{m}$ is exactly half of $340\ \text{m/s}$, so the round trip takes precisely one second. Recognising that saves the division entirely.

4. The same method in other places

Ultrasound scanning, sonar depth-sounding and ultrasonic parking sensors all use the identical relation $s = \dfrac{vt}{2}$. Whenever a wave is detected back where it started, the factor of two is there.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $0.25\ \text{s}$ — Halved Twice
  A quarter of the round trip - the distance halved as well as the return journey omitted.
• B. $0.5\ \text{s}$ — One-Way Only
  The one-way time, $\dfrac{170}{340}$. The echo has to come back as well.
• D. $2.0\ \text{s}$ — Doubled Twice
  Twice the answer, from $4 \times 170$ - the round trip counted twice over.
• E. $0.2\ \text{s}$ — Not Consistent
  Corresponds to a path of $68\ \text{m}$, which matches neither the single nor the double journey.

Common Mistake (⚠️):
Using the one-way distance and getting $0.5\ \text{s}$. An echo is heard back at the source, so the sound has made the journey twice.

Takeaway (📌):
Echo problems: double the distance, then divide by the speed. The factor of two is the whole question.

Correct Answer: D. $+2\ \text{nC}$ on each

Fastest Approach (🚀):
Total: $+6 + (-2) = +4\ \text{nC}$.
Identical spheres share equally: $\dfrac{+4}{2} = +2\ \text{nC}$ each.

Matches Option D.

Step-by-Step Breakdown:

1. Charge is conserved

Touching the spheres lets electrons move between them, but none are created or destroyed. The total charge is therefore the same before and after:
$q_{\text{total}} = (+6) + (-2) = +4\ \text{nC}$

Adding the charges as signed numbers is the step that matters; treating the $-2$ as a magnitude gives $+8$ and everything after it is wrong.

2. Identical conductors share equally

While in contact the two spheres form one conductor. Because they are the same size and shape, the charge distributes symmetrically, so on separation each carries half:
$q = \dfrac{+4}{2} = +2\ \text{nC}$

3. What physically moved

Electrons flowed from the negative sphere to the positive one until the two reached the same potential. In this case $4\ \text{nC}$ of negative charge moved across - about $2.5\times10^{10}$ electrons.

4. Why 'identical' is in the question

Spheres of different sizes would still conserve charge but would not share it equally: the larger one takes more. The equal split is a consequence of the symmetry, not of the contact alone.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $+6\ \text{nC}$ and $-2\ \text{nC}$, unchanged — No Transfer
  Contact between conductors allows electrons to move, so the charges cannot stay as they were.
• B. $+4\ \text{nC}$ on each — Signs Ignored
  The signs ignored: $\dfrac{6+2}{2} = 4$. Adding as signed numbers gives a total of $+4$, and each sphere takes half of that.
• C. Zero on each — Full Cancellation
  Complete cancellation would need equal and opposite charges. Here $+6$ and $-2$ leave a surplus of $+4$.
• E. $-2\ \text{nC}$ on each — Charge Copied
  One sphere's original charge given to both. Charge is shared, not copied.

Common Mistake (⚠️):
Ignoring the sign and averaging $6$ and $2$ to get $+4\ \text{nC}$ each. Charges add as signed quantities, so the opposite charges partly cancel.

Takeaway (📌):
Total first, using signs; then divide by two if - and only if - the conductors are identical.

Correct Answer: C. $20\ \text{cm}$

Fastest Approach (🚀):
Anticlockwise: $2.0 \times 30 = 60$.
Clockwise: $3.0 \times d = 60 \implies d = 20\ \text{cm}$.

Matches Option C.

Step-by-Step Breakdown:

1. Why the rule's own weight can be ignored

The rule is uniform and pivoted at its centre, so its weight acts through the pivot and has zero moment about it. Only the two hanging weights matter.

2. The principle of moments

For equilibrium, the total clockwise moment about the pivot equals the total anticlockwise moment:
$F_1 d_1 = F_2 d_2$

3. Substitute and solve

$2.0 \times 30 = 3.0 \times d$

$60 = 3.0d \implies d = 20\ \text{cm}$

4. Two checks worth doing

First, the heavier weight sits closer to the pivot, which is what a see-saw does. Second, the distances stayed in centimetres on both sides, so no conversion is needed - the units cancel in the equality.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $45\ \text{cm}$ — Fails the Check
  Would need $3.0 \times 45 = 135$ against the left-hand $60$ - more than twice as much.
• B. $30\ \text{cm}$ — Distance Copied
  Copies the left-hand distance, which would only balance if the two weights were equal.
• D. $15\ \text{cm}$ — Wrong Ratio
  Halves the left-hand distance, which corresponds to a $4.0\ \text{N}$ weight rather than $3.0\ \text{N}$.
• E. $10\ \text{cm}$ — Fails the Check
  Gives a clockwise moment of $30$, only half the anticlockwise $60$.

Common Mistake (⚠️):
Converting one distance to metres and not the other. Because moments are compared, any consistent unit works - but it must be the same on both sides.

Takeaway (📌):
Moments balance as force times distance. The heavier weight always ends up nearer the pivot.

Correct Answer: E. Each piece is a complete magnet, with its own north and south pole

Fastest Approach (🚀):
Every domain is itself a dipole $\Rightarrow$ a pole cannot be isolated.
Each half is a complete, weaker magnet.

Matches Option E.

Step-by-Step Breakdown:

1. Where the magnetism actually resides

A bar magnet is not 'north at one end and south at the other' in the way a charged rod is positive at one end. Its magnetism comes from domains throughout the material, all aligned the same way, and every domain is itself a north-south pair.

2. What cutting it does

Cutting between the poles does not separate two kinds of thing; it just gives two shorter rows of aligned domains. New poles appear immediately at the cut faces, so each half is a complete magnet - weaker, but with both poles.

3. Keep cutting

The argument does not stop. Halve each piece again and again and every fragment is still a dipole, right down to the individual atom. An isolated magnetic pole - a monopole - has never been observed.

4. How the halves behave

Bring the two halves back together the way they were cut and they attract, because the new south face of one meets the new north face of the other. Turn one round and they repel.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. One piece is a north pole only and the other is a south pole only — Monopole Assumed
  The magnetic monopole, which does not exist. New poles form at the cut faces the instant the magnet is divided.
• B. Both pieces lose their magnetism entirely — Magnetism Lost
  Cutting does not disturb the alignment of the domains; heating or hammering the magnet would, but a clean cut does not.
• C. Each piece has two north poles — Poles Duplicated
  Every magnet has one of each. A piece with two north poles would have field lines with nowhere to return to.
• D. The two pieces repel each other, because both are now north — Poles Duplicated
  Rests on the same impossibility as C, and in fact the halves attract when brought back as they were cut.

Common Mistake (⚠️):
Reasoning by analogy with electric charge, where positive and negative really can be separated. Magnetic poles cannot: there is no magnetic equivalent of an isolated charge.

Takeaway (📌):
Poles always come in pairs. Cutting a magnet makes two smaller magnets, never two isolated poles.

Correct Answer: B. Mass $80\ \text{kg}$, weight $128\ \text{N}$

Fastest Approach (🚀):
Mass unchanged: $80\ \text{kg}$.
Weight $= mg = 80 \times 1.6 = 128\ \text{N}$.

Matches Option B.

Step-by-Step Breakdown:

1. Mass does not travel-change

Mass measures how much matter there is, and equivalently how hard the body is to accelerate. Neither depends on where the body happens to be, so the astronaut is $80\ \text{kg}$ on the Moon, in orbit, or anywhere else. Its unit is the kilogram.

2. Weight does

Weight is the gravitational force on the body, $W = mg$, measured in newtons. It scales directly with the local field strength:
$W = 80 \times 1.6 = 128\ \text{N}$

3. Compare with the Earth value

On Earth the same astronaut weighs $80 \times 10 = 800\ \text{N}$. The Moon figure is $\dfrac{1.6}{10} = 0.16$ of that, which is the familiar 'about a sixth of Earth gravity'.

4. Spotting the wrong answers by unit alone

Any option quoting a mass in newtons or a weight in kilograms is wrong before the numbers are read. Here the trap is subtler - the numbers are swapped between the two quantities - so check which number carries which unit.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. Mass $12.8\ \text{kg}$, weight $128\ \text{N}$ — Mass Scaled
  Right weight, but the mass has been scaled by the field strength too. Mass does not depend on location.
• C. Mass $80\ \text{kg}$, weight $800\ \text{N}$ — Earth Weight
  Right mass, but the Earth's weight quoted. The question asks for the weight on the Moon.
• D. Mass $12.8\ \text{kg}$, weight $800\ \text{N}$ — Both Wrong
  Both halves wrong: mass scaled and the Earth weight retained.
• E. Mass $80\ \text{kg}$, weight $80\ \text{N}$ — Wrong g
  A weight of $80\ \text{N}$ would need $g = 1\ \text{N/kg}$. The Moon's value is $1.6$.

Common Mistake (⚠️):
Scaling the mass by the gravity ratio. Mass is unaffected by the field; it is the weight that changes.

Takeaway (📌):
Mass in kilograms is invariant. Weight in newtons is $mg$ and follows the local field.

Correct Answer: A. A deflection while the magnet is moving in, falling to zero once it is held still

Fastest Approach (🚀):
Moving in $\Rightarrow$ flux changing $\Rightarrow$ e.m.f. induced $\Rightarrow$ current.
Held still $\Rightarrow$ flux constant $\Rightarrow$ no e.m.f. $\Rightarrow$ zero.

Matches Option A.

Step-by-Step Breakdown:

1. What induction actually responds to

A voltage is induced when the magnetic flux linking the coil changes. The size of the induced e.m.f. depends on how fast it changes, not on how large the field is.

2. While the magnet moves in

More and more field lines thread the coil, so the flux is increasing. An e.m.f. is induced and, because the circuit is complete through the ammeter, a current flows and the needle deflects.

3. Once the magnet is held still

The field inside the coil is now strong but constant. Nothing is changing, so no e.m.f. is induced and the needle returns to zero - even though the magnet is still there.

4. And on the way out

Pulling the magnet out reduces the flux, inducing an e.m.f. again but in the opposite sense, so the needle deflects the other way. Lenz's law explains the sense: the induced current always opposes the change that produced it, so the coil resists the magnet coming in and resists it leaving.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. A steady deflection for the whole time the magnet is inside the coil — Flux Not Rate
  Would require a constant field to induce a constant e.m.f. Induction responds to the rate of change, which is zero once the magnet stops.
• C. No deflection at any point — Induction Denied
  Moving the magnet in genuinely does change the flux, so there is certainly a deflection during that stage.
• D. A deflection that grows larger while the magnet is held still — Flux Not Rate
  Nothing is changing while it is held still, so there is nothing to grow.
• E. A deflection only once the magnet has stopped moving — Reversed
  Exactly inverted: the deflection occurs while it moves and stops when it stops.

Common Mistake (⚠️):
Thinking the field inside the coil is what drives the current. A stationary magnet inside a coil induces nothing at all, no matter how strong it is.

Takeaway (📌):
Induction needs a rate of change of flux. Movement produces a reading; a stationary magnet, however strong, produces none.