ESAT Paper 3 Physics Worked Solutions

Correct Answer: C. $0.5\text{ m s}^{-2}$ accelerating

Fastest Approach (🚀):
$a = \frac{500 - 460}{80} = 0.5\text{ms}^{-2}$ accelerating.
Matches Option C.

Step-by-Step Breakdown:

1. Calculating Net Force

The forces acting along the incline are:

• Downward force along slope: $F_{\text{down}} = 500\text{N}$
• Resistive air force: $F_{\text{resist}} = 460\text{N}$

Since $F_{\text{down}} > F_{\text{resist}}$, the net force acts down the slope:
$F_{\text{net}} = F_{\text{down}} - F_{\text{resist}} = 500 - 460 = 40\text{N}$

2. Applying Newton's Second Law

Using $F_{\text{net}} = m a$ with mass $m = 80\ \text{kg}$:
$a = \frac{F_{\text{net}}}{m} = \frac{40\text{N}}{80\ \text{kg}} = 0.5\text{ms}^{-2}$

3. Determining Motion State & Option Matching

Since net force is in the direction of downward motion, the skier accelerates at $0.5\text{ms}^{-2}$.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $1.0\text{ m s}^{-2}$ accelerating — Arithmetic Error
  The direction is right, but $1.0\text{ m s}^{-2}$ needs a net force of $80\text{ N}$. The net force here is $500 - 460 = 40\text{ N}$, and $40/80 = 0.5$.
• B. $0.5\text{ m s}^{-2}$ decelerating — Direction Error
  Right magnitude, wrong direction. The $500\text{ N}$ down the slope is larger than the $460\text{ N}$ resisting it, so the net force acts down the slope and the skier speeds up.
• D. $1.0\text{ m s}^{-2}$ decelerating — Direction and Magnitude
  Both errors at once: the magnitude would need an $80\text{ N}$ net force, and the net force acts down the slope, so this is an acceleration.
• E. $0.3\text{ m s}^{-2}$ accelerating — Arithmetic Error
  $0.3\text{ m s}^{-2}$ would need a net force of $24\text{ N}$. Subtracting the two given forces gives $500 - 460 = 40\text{ N}$.

Common Mistake (⚠️):
Adding forces $500 + 460 = 960\text{N}$ instead of subtracting resistive air drag.

Takeaway (📌):
Resistive forces (friction, air resistance) always oppose motion.

Correct Answer: C. $\frac{1}{\sqrt{2}}v$

Fastest Approach (🚀):
$v^2 \propto F \implies (v')^2 = v^2/2 \implies v' = v/\sqrt{2}$.
Matches Option C.

Step-by-Step Breakdown:

1. Work-Energy Theorem for Braking

Assuming the stopping distance $s$ is entirely braking distance (negligible reaction time), the work done by the brakes equals the initial kinetic energy of the cyclist:
$F \cdot s = \frac{1}{2}mv^2$
$s = \frac{mv^2}{2F}$

2. Analysing Rainy Conditions

In the rain, the braking force is halved to $F/2$. Let the new velocity be $v'$. The new stopping distance is:
$s' = \frac{m(v')^2}{2(F/2)} = \frac{m(v')^2}{F}$

3. Equating Stopping Distances

We want the new stopping distance to equal the usual stopping distance ($s' = s$):
$\frac{m(v')^2}{F} = \frac{mv^2}{2F}$
$(v')^2 = \frac{v^2}{2}$
$v' = \frac{v}{\sqrt{2}}$

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $\frac{1}{2}v$ — Speed Scaling Error
  Treating stopping distance as proportional to $v$ rather than $v^2$. The work equation $Fs = \tfrac12 mv^2$ squares the speed, so halving the force halves $v^2$ and the speed falls by a factor of $\sqrt{2}$, not $2$.
• B. $\frac{1}{4}v$ — Double Application
  The halving applied twice - once to $v^2$ and again to $v$. One application is enough: $\tfrac{F}{2}s = \tfrac12 m v'^2$ gives $v'^2 = \tfrac{v^2}{2}$.
• D. $\frac{1}{3}v$ — Wrong Factor
  A factor of $3$ appears from nowhere. The rain halves the braking force, it does not reduce it to a third.
• E. $\frac{1}{\sqrt{3}}v$ — Wrong Factor
  The square root is right, but the factor is not: the force is halved, so $v' = \tfrac{v}{\sqrt{2}}$, not $\tfrac{v}{\sqrt{3}}$.

Common Mistake (⚠️):
Thinking $v \propto F$ directly and concluding the velocity must be halved (Option A). Kinetic energy depends on $v^2$.

Takeaway (📌):
Stopping distance (ignoring reaction time) is proportional to the square of the velocity and inversely proportional to the braking force.

Correct Answer: E. $\frac{M_1 v_1 + M_2 v_2}{M_1 + M_2}$

Fastest Approach (🚀):
$M_1 v_1 + M_2 v_2 = (M_1 + M_2)V \implies V = \frac{M_1 v_1 + M_2 v_2}{M_1 + M_2}$.
Matches Option E.

Step-by-Step Breakdown:

1. Applying Conservation of Momentum

The law of conservation of momentum states that the total momentum before the collision equals the total momentum after the collision:
$P_{\text{initial}} = P_{\text{final}}$

2. Formulating Momenta

• Total initial momentum is the sum of the individual momenta: $P_{\text{initial}} = M_1 v_1 + M_2 v_2$.
• Since the particles stick together, they form a single combined mass $(M_1 + M_2)$ moving at a final velocity $V$.
• Total final momentum: $P_{\text{final}} = (M_1 + M_2)V$.

3. Calculating Final Velocity

Equating the initial and final momenta:
$M_1 v_1 + M_2 v_2 = (M_1 + M_2)V$
$V = \frac{M_1 v_1 + M_2 v_2}{M_1 + M_2}$

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\frac{M_1 v_1 - M_2 v_2}{M_1 - M_2}$ — Sign and Denominator
  Both parts wrong. The two momenta add to give the system's total, and the particles stick together, so the mass afterwards is $M_1 + M_2$. Mass is never lost in a collision.
• B. $\frac{M_1 v_1 + 2M_2 v_2}{M_1 + M_2}$ — Spurious Factor
  A spurious factor of $2$ on $M_2v_2$. Particle 2's momentum is simply $M_2v_2$.
• C. $\frac{M_1 v_1 + M_2 v_2}{M_1 - M_2}$ — Denominator Error
  The numerator is right, but the particles carry on stuck together, so the moving mass afterwards is $M_1 + M_2$, not $M_1 - M_2$.
• D. $\frac{M_1 v_1 - M_2 v_2}{M_1 + M_2}$ — Sign Error
  Subtracting the momenta. Both are momenta of the same system and add; a minus sign would only appear if $v_2$ were measured in the opposite direction, which the question does not state.

Common Mistake (⚠️):
Using a minus sign in the numerator assuming the velocities must be in opposite directions. The variables $v_1$ and $v_2$ represent vector velocities (can be negative intrinsically).

Takeaway (📌):
For perfectly inelastic collisions, $V = \frac{\sum p_i}{\sum m_i}$.

Correct Answer: F. 20 m

Fastest Approach (🚀):
$1000 \times d = \frac{1}{2}(10000)(2.0)^2 = 20000 \implies d = 20\text{m}$.
Matches Option F.

Step-by-Step Breakdown:

1. Applying the Work-Energy Theorem

The kinetic energy of the spaceship must be entirely dissipated by the work done by the robotic arm.
$W = \Delta KE$
$F \cdot d = \frac{1}{2} m v^2$

2. Calculating the Stopping Distance

Given $F = 1000\text{N}$, $m = 10,000\ \text{kg}$, and $v = 2.0\text{ms}^{-1}$:
$1000 \cdot d = \frac{1}{2} (10000) (2.0)^2$
$1000 \cdot d = 5000 \times 4$
$1000 \cdot d = 20000$
$d = 20\text{m}$

Matches Option F.

Why the Other Options Are Wrong (❌):

• A. 0.050 m — Order of Magnitude
  The kinetic energy is $\tfrac12(10\,000)(2.0)^2 = 20\,000\text{ J}$, so $d = 20\,000/1000 = 20\text{ m}$. This is $400$ times smaller - a decimal slip.
• B. 0.10 m — Order of Magnitude
  $200$ times too small. $Fd = \tfrac12 mv^2$ gives $d = 20\,000/1000 = 20\text{ m}$.
• C. 0.20 m — Order of Magnitude
  $100$ times too small: this is $20\,000$ divided by $100\,000$ rather than by the $1000\text{ N}$ force.
• D. 5.0 m — Wrong Value Used
  Uses $v = 1.0\text{ m s}^{-1}$ instead of $2.0$: $\tfrac12(10\,000)(1.0)^2/1000 = 5.0\text{ m}$.
• E. 10 m — Missing Square
  The square has been dropped: $\tfrac12(10\,000)(2.0)/1000 = 10\text{ m}$. Kinetic energy goes as $v^2$.

Common Mistake (⚠️):
Calculating acceleration $a = F/m = 0.1\text{ms}^{-2}$ but making a mistake in $v^2 = u^2 + 2as$ by dropping a factor of 2.

Takeaway (📌):
Work-energy theorem is often faster than calculating intermediate acceleration and time for stopping distance problems.

Correct Answer: B. 43200 cm

Fastest Approach (🚀):
1 cycle = $4(6.0) = 24.0\ \text{cm}$. 1 min = $60 \times 30 = 1800$ cycles. Total distance = $1800 \times 24.0 = 43200\ \text{cm}$. Matches Option B.

Step-by-Step Breakdown:

1. Understanding Particle Motion

In a transverse wave, particles of the medium oscillate up and down in simple harmonic motion (SHM). They do not travel along with the wave.
In one complete wave cycle, a particle moves from equilibrium to a crest (distance $A$), back to equilibrium (distance $A$), to a trough (distance $A$), and back to equilibrium (distance $A$).
Thus, the total distance traveled by a single particle in one cycle is $4\ \text{A}$.
$D_{\text{cycle}} = 4 \times 6.0\ \text{cm} = 24.0\ \text{cm}$

2. Calculating Total Cycles

The frequency is $f = 30.0\text{Hz}$, meaning 30 cycles occur per second.
In one minute ($t = 60\text{s}$), the total number of cycles is:
$N = f \times t = 30.0 \times 60 = 1800\ \text{cycles}$

3. Calculating Total Distance

$\text{Total distance} = N \times D_{\text{cycle}}$
$\text{Total distance} = 1800 \times 24.0\ \text{cm} = 43200\ \text{cm}$

*(Note: The wavelength $0.8\ \text{cm}$ is distractor information and is not needed for this calculation.)*

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. 880 cm — Time Conversion
  In one cycle a particle covers $4A = 24\text{ cm}$, and the wave completes $30 \times 60 = 1800$ cycles in a minute, so the distance is $1800 \times 24 = 43\,200\text{ cm}$. $880\text{ cm}$ is about $37$ cycles' worth.
• C. 864 cm — Time Conversion
  $864\text{ cm}$ is exactly $36$ cycles of $4A$. The wave completes $30 \times 60 = 1800$ cycles in a minute, not $36$ - the frequency has not been multiplied by the $60$ seconds.
• D. 748 cm — Method Error
  Neither $4A \times 1800$ nor either of the usual partial answers ($A \times 1800 = 10\,800$, $2A \times 1800 = 21\,600$). A particle covers $4A = 24\text{ cm}$ per cycle and there are $1800$ cycles.
• E. 2440 cm — Time Conversion
  About $102$ cycles' worth. The particle completes $1800$ cycles in the minute, each covering $4A = 24\text{ cm}$.
• F. 1680 cm — Time Conversion
  $70$ cycles' worth of $4A$. The count of cycles is $30\text{ Hz} \times 60\text{ s} = 1800$.

Common Mistake (⚠️):
Calculating the distance the wave travels ($v = f \lambda$) instead of the distance $a$ particle travels, or using $2\ \text{A}$ per cycle.

Takeaway (📌):
In transverse waves, local particles travel $4\ \text{A}$ per cycle, completely independent of the wave's horizontal propagation speed or wavelength.

Correct Answer: E. 6 days

Fastest Approach (🚀):
$2 (1/2)^{t/2} = (1/2)^{t/3} \implies 2 = (1/2)^{t/3 - t/2} = (1/2)^{-t/6} = 2^{t/6} \implies t = 6$. Matches Option E.

Step-by-Step Breakdown:

1. Setting Up the Decay Equations

The number of remaining atoms after time $t$ is given by $N(t) = N_0 (1/2)^{t/t_{1/2}}$.

• For isotope X: $N_X(t) = N_{X0} (1/2)^{t/2}$
• For isotope Y: $N_Y(t) = N_{Y0} (1/2)^{t/3}$

We are given that initially, $N_{X0} = 2 N_{Y0}$.

2. Equating the Atoms

We want to find $t$ when $N_X(t) = N_Y(t)$:
$2 N_{Y0} \left(\frac{1}{2}\right)^{t/2} = N_{Y0} \left(\frac{1}{2}\right)^{t/3}$

Divide both sides by $N_{Y0}$:
$2 \left(\frac{1}{2}\right)^{t/2} = \left(\frac{1}{2}\right)^{t/3}$

3. Solving for $t$

Isolate the base $1/2$ terms:
$2 = \frac{(1/2)^{t/3}}{(1/2)^{t/2}} = \left(\frac{1}{2}\right)^{t/3 - t/2}$

Simplify the exponent:
$\frac{t}{3} - \frac{t}{2} = \frac{2t - 3t}{6} = -\frac{t}{6}$

So we have:
$2 = \left(\frac{1}{2}\right)^{-t/6} = 2^{t/6}$

Equating exponents:
$1 = \frac{t}{6} \implies t = 6\ \text{days}$

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. The expected number of atoms of X and Y is never equal. — Wrong Conclusion
  They do become equal. X starts with twice as many atoms but decays faster, so Y catches it: $2(\tfrac12)^{t/2} = (\tfrac12)^{t/3}$ gives $t = 6$ days.
• B. 2 days — Misread Quantity
  $2$ days is X's half-life, not the crossing time. At $t = 2$, X has fallen to $1.00$ and Y to $0.63$ (in units of Y's initial count) - X is still ahead.
• C. 3 days — Misread Quantity
  $3$ days is Y's half-life. At $t = 3$, X is at $0.71$ and Y at $0.50$, so X is still ahead.
• D. 4 days — Premature Stop
  At $t = 4$, X is at $0.50$ and Y at $0.40$ - close, but X is still ahead. The equality is at $t = 6$ days, where both are at $0.25$.
• F. 12 days — Exponent Error
  Twice the answer. The exponents differ by $\tfrac{t}{2} - \tfrac{t}{3} = \tfrac{t}{6}$, so $2 = 2^{t/6}$ and $t = 6$; writing $\tfrac{t}{12}$ doubles it.

Common Mistake (⚠️):
Setting up the initial condition backwards (e.g. $N_{Y0} = 2N_{X0}$), which would lead to negative time or no solution.

Takeaway (📌):
When comparing two exponential decays, convert all terms to the same base (e.g., base 2) and equate the exponents.

Correct Answer: E. 0.1 Hz

Fastest Approach (🚀):
$T = 2 \times 5 = 10\text{s} \implies f = 0.1\text{Hz}$.
Matches Option E.

Step-by-Step Breakdown:

1. Identifying Wave Cycle Phase

In a longitudinal wave, compression and rarefaction are half a wavelength (and half a time period $T/2$) apart.
$\frac{T}{2} = 5\text{s} \implies T = 10\text{s}$

2. Calculating Frequency

$f = \frac{1}{T} = \frac{1}{10\text{s}} = 0.1\text{Hz}$

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. 0.5 Hz — Period Error
  $0.5\text{ Hz}$ is a period of $2\text{ s}$. Rarefaction to compression is half a cycle, so $T = 2 \times 5 = 10\text{ s}$ and $f = 0.1\text{ Hz}$.
• B. 0.4 Hz — Period Error
  $0.4\text{ Hz}$ is a period of $2.5\text{ s}$, which does not follow from a $5\text{ s}$ half-cycle.
• C. 0.3 Hz — Period Error
  $0.3\text{ Hz}$ is a period of about $3.3\text{ s}$. The half-cycle is $5\text{ s}$, so the period is $10\text{ s}$.
• D. 0.2 Hz — Half-Cycle Missed
  Treating the $5\text{ s}$ as a whole period: $1/5 = 0.2\text{ Hz}$. Rarefaction to compression is only half a cycle.

Common Mistake (⚠️):
Assuming $5\text{s}$ is a full period, getting $f = 0.2\text{Hz}$.

Takeaway (📌):
Peak-to-trough (or compression-to-rarefaction) time is ALWAYS half the period $T/2$.

Correct Answer: B. B, W

Fastest Approach (🚀):
Sparse lines $\Rightarrow$ low density $\Rightarrow$ rarefaction $\Rightarrow$ B on the strip.
The interval lying wholly under a trough $\Rightarrow$ W on the trace.

Matches Option B.

Step-by-Step Breakdown:

1. What a rarefaction is

Sound is a longitudinal wave, so the medium alternately bunches up and spreads out. A compression is where the particles crowd together; a rarefaction is the opposite - particles spread apart, density and pressure below their resting values. Density and pressure move together, which is what lets the two diagrams be read against each other.

2. Reading the particle strip

• A sits beneath the tightly packed band at the left-hand end. High density, so this is a compression.
• B sits beneath the widely spaced region - the lines there are roughly two and a half times further apart than around A. Low density, so this is the rarefaction.

3. Reading the sine trace

The four coloured bars each mark an interval of time:

• V spans the first crest - a maximum, so a compression.
• Y starts in the trough's tail and ends on the second crest: it straddles the upward zero crossing.

X starts on the falling part of the first crest and ends at the trough: it straddles the downward zero crossing rather than sitting in one phase.
W lies wholly within the below-axis half-cycle - the trough, the minimum, the rarefaction.

The straddling intervals X and Y are the trap: each touches the trough without representing it.

4. Putting the two together

The rarefaction is the sparse region B on the strip and the trough interval W on the trace.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. B, X — Straddles Crossing
  B is the right region on the strip, but X straddles the downward zero crossing: most of that interval is still above the axis, so it does not represent the trough.
• C. A, X — Compression Chosen
  Wrong on both counts: A sits beneath the densely packed band, a compression, and X straddles a zero crossing rather than spanning the trough.
• D. A, Y — Compression Chosen
  A is a compression, and Y runs from the trough's tail up onto the second crest - it straddles the upward crossing instead of spanning the minimum.
• E. A, W — Strip Misread
  W is the right interval, but A sits beneath the tightly packed lines - the compression. The sparse region is B.

Common Mistake (⚠️):
Choosing an interval that merely touches the trough. X ends at the trough and Y begins in its tail, but each straddles a zero crossing, so most of each interval is not rarefaction at all. Only W lies wholly within the below-axis half-cycle.

Takeaway (📌):
Rarefaction is low density and low pressure together: the spread-out region of a particle diagram, and the interval spanning - not merely touching - a trough.

Correct Answer: D. A slightly permeable, conducting fabric

Fastest Approach (🚀):
$V \propto nT$. To keep $V$ small, we want $n$ to drop (permeable) and $T$ to drop (conducting). Matches Option D.

Step-by-Step Breakdown:

1. Understanding the Bursting Mechanism

The balloon bursts when its volume $V$ becomes too large.
According to the ideal gas law, $V = \frac{nRT}{P}$.
As the balloon climbs, external pressure $P$ decreases, which causes $V$ to increase.

2. Minimizing Volume Expansion

To maximize the climbing altitude before bursting, we must counteract the increase in $V$ by decreasing the numerator $nT$.

• Temperature ($T$): The atmosphere gets colder with altitude. If the balloon fabric is conducting, heat transfers out of the balloon, lowering the internal temperature $T$.
• Amount of Gas ($n$): If the fabric is slightly permeable, some gas escapes over time, reducing the number of moles $n$.

Since the prompt guarantees the balloon will still have enough gas to provide buoyancy, these two properties (conducting and permeable) will slow the volume expansion, allowing the balloon to climb higher.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. An impermeable, insulating fabric — Both Properties Wrong
  Both properties work the wrong way. An insulating fabric keeps the gas warm while the air outside cools, and an impermeable one traps every molecule, so $V$ grows as fast as possible and the balloon bursts soonest.
• B. An impermeable, conducting fabric — One Property Wrong
  Conducting is right - letting the gas cool with the atmosphere shrinks it - but sealing the envelope holds $n$ fixed, so the balloon still expands more than it needs to as the outside pressure falls.
• C. A slightly permeable, insulating fabric — One Property Wrong
  Permeable is right - losing gas lowers $n$ - but an insulating fabric keeps $T$ high while the surrounding air cools, which works against you.
• E. None of the above — Premature Rejection
  Option D does work. From $pV = nRT$, a conducting fabric lowers $T$ and a permeable one lowers $n$, and both slow the growth of $V$.

Common Mistake (⚠️):
Thinking we need an insulating, impermeable balloon to 'protect' the gas. This would keep $T$ and $n$ high, making it burst sooner.

Takeaway (📌):
Counteracting pressure drops with temperature drops and gas loss keeps balloon volume from expanding past its bursting point.

Correct Answer: B. $\frac{1}{2}\text{W}$

Fastest Approach (🚀):
$P_1 = V^2/R_1 \implies 1 = V^2/2 \implies V^2 = 2$. $P_2 = V^2/R_2 = 2/4 = 1/2\text{W}$. Matches Option B.

Step-by-Step Breakdown:

1. Identifying Shared Voltage

In a parallel circuit, the voltage $V$ across each branch is identical.

2. Using the Power Equation

The power dissipated by a resistor is $P = \frac{V^2}{R}$.
For resistor $R_1$ ($2\ \text{\Omega}$):
$P_1 = \frac{V^2}{R_1} \implies 1\text{W} = \frac{V^2}{2\Omega} \implies V^2 = 2\text{V}^2$

3. Calculating Power for $R_2$

For resistor $R_2$ ($4\ \text{\Omega}$):
$P_2 = \frac{V^2}{R_2} = \frac{2}{4} = \frac{1}{2}\text{W}$

Alternatively, since $P \propto \frac{1}{R}$ for constant $V$, doubling the resistance halves the power. $4\ \text{\Omega}$ is twice $2\ \text{\Omega}$, so power is $1\text{W} / 2 = 1/2\text{W}$.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. 8 W — Inverted Relationship
  Scaling the wrong way. The three resistors share the same voltage, so $P = V^2/R$ means more resistance gives less power, not more.
• C. $\frac{4}{\sqrt{2}}\text{W}$ — Spurious Root
  There is no square root in $P = V^2/R$. The relationship is linear in $1/R$, so doubling $R$ simply halves $P$.
• D. $\frac{1}{4}\text{W}$ — Wrong Power
  Squaring the resistance ratio. $P \propto 1/R$, not $1/R^2$, so the factor is $2$, not $4$.
• E. $\frac{7}{4}\text{W}$ — Irrelevant Term
  $R_3$ does not enter the comparison at all: in parallel each branch sees the same voltage, so $P_2 = P_1 \times R_1/R_2 = \tfrac12\text{ W}$.

Common Mistake (⚠️):
Using $P = I^2 R$ and assuming current is the same in parallel branches, concluding that power is proportional to $R$.

Takeaway (📌):
In parallel circuits, use $P = V^2/R$ to compare power since $V$ is constant across branches.

Correct Answer: A. The other magnet will turn clockwise

Fastest Approach (🚀):
Magnetic dipoles coupled together act like gears. Anticlockwise $\implies$ Clockwise. Matches Option A.

Step-by-Step Breakdown:

1. Initial Alignment

When two bar magnets are placed near each other on pivots, they align to minimize their magnetic potential energy. This means the North pole of one magnet points towards the South pole of the other.

2. Analysing the Rotation

Suppose the magnets are side-by-side, with Magnet 1's North pole facing Magnet 2's South pole.
If Magnet 1 is slowly rotated anticlockwise, its North pole moves in an arc (e.g., upwards).
Because unlike poles attract, Magnet 2's South pole is pulled along the same path (upwards) to stay as close to Magnet 1's North pole as possible.

3. Resulting Motion

For the adjacent pole of Magnet 2 to follow the arc of Magnet 1's pole, Magnet 2 must rotate in the opposite sense.
An upward movement on its adjacent side corresponds to a clockwise rotation.
Thus, two coupled bar magnets behave exactly like a pair of mechanical gears.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. The other magnet will turn anticlockwise — Gear Sense
  Treats the pair as if one dragged the other like a rigid body. Coupled dipoles behave like meshed gears: the facing poles attract, so the driven magnet turns the opposite way.
• C. The other magnet will oscillate back and forth — Steady Drive
  Oscillation needs the driving pole to reverse. It does not - the first magnet turns steadily in one direction, so the second follows steadily.
• D. The other magnet turns 180 degrees, and remains in that position — Drive Stops
  Would need the driving field to stop moving. It keeps rotating, so the second magnet keeps being pulled round.
• E. The other magnet turns 90 degrees, then remains in that position. — Unstable Rest
  Same objection, and $90^\circ$ is the least stable alignment for two dipoles - it is where the torque between them is greatest, not where one comes to rest.

Common Mistake (⚠️):
Thinking they are 'dragged' in the same direction (anticlockwise) like a solid object, ignoring the pivot mechanics.

Takeaway (📌):
Coupled pivoted magnets rotate in opposite directions, analogous to intermeshing gears.

Correct Answer: B. $\frac{50}{3}\text{m}$

Fastest Approach (🚀):
$W = \frac{1}{2} m v^2 = \frac{1}{2}(4)(100) = 200\text{J}$.
$W = F \times d \implies 200 = 12d \implies d = \frac{200}{12} = \frac{50}{3}\text{m}$.
Matches Option B.

Step-by-Step Breakdown:

1. Work-Energy Theorem

The work done to bring a moving body to rest equals its initial kinetic energy:
$W = \Delta KE = \frac{1}{2} m v^2$
$W = \frac{1}{2} \times 4\ \text{kg} \times (10\text{ms}^{-1})^2 = 2 \times 100 = 200\text{J}$

2. Calculating Distance

Work is also defined as Force times distance:
$W = F \times d$
$200 = 12 \times d$
$d = \frac{200}{12} = \frac{50}{3}\text{m}$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\frac{\sqrt{50}}{3}\text{m}$ — Spurious Root
  A square root appears from nowhere. Work done is force $\times$ distance, so $d = 200/12 = \tfrac{50}{3}\text{ m}$.
• C. $\frac{10}{3}\text{m}$ — Wrong Quantity
  This is $40/12$ - using the momentum $mv = 40$ in place of the kinetic energy $\tfrac12 mv^2 = 200\text{ J}$.
• D. $\frac{100}{3}\text{m}$ — Missing Half
  This is $400/12$ - the factor $\tfrac12$ has been dropped from $\tfrac12 mv^2$.
• E. $\frac{30}{3}\text{m}$ — Arithmetic Error
  $\tfrac{30}{3} = 10\text{ m}$. The work-energy calculation gives $200\text{ J} \div 12\text{ N} = \tfrac{50}{3} \approx 16.7\text{ m}$.

Common Mistake (⚠️):
Forgetting the $1/2$ factor in kinetic energy, obtaining $400\text{J}$.

Takeaway (📌):
Work required to stop a body $= \frac{1}{2} m v^2$.

Correct Answer: B. 10 m

Fastest Approach (🚀):
$1,000,000 = 100,000 \times d \implies d = 10\text{m}$. Matches Option B.

Step-by-Step Breakdown:

1. Identifying the Moment Equation

The moment (torque) $\tau$ around the base of the crane is given by the force (weight of the mass) multiplied by the perpendicular horizontal distance $d$ from the base:
$\tau = F \times d$

2. Setting up the Inequality

To prevent toppling, the moment must not exceed $1\ \text{mega-Newton metre}$ ($1,000,000\text{Nm}$):
$F \times d \le 1,000,000\text{Nm}$

Given the weight $F = 100,000\text{N}$:
$100,000 \times d \le 1,000,000$

3. Solving for Distance

$d \le \frac{1,000,000}{100,000} = 10\text{m}$

Thus, the maximum horizontal distance is $10\text{m}$. (Note: The height of the crane and the vertical lifting distance are irrelevant to the moment at the base).

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. 1 m — Load Misread
  Needs a load of $10^{6}\ \text{N}$. The load is $10^{5}\ \text{N}$, so the limiting distance is $\tfrac{10^{6}}{10^{5}}=10\ \text{m}$.
• C. 100 m — Figure Copied
  The arm's length, copied from the question. The crane would topple long before the load reached the end of the arm.
• D. $\frac{100}{3}\text{m}$ — Height Used
  Brings the $30\ \text{m}$ lift height into the moment. A moment about the base uses the horizontal distance only - lifting straight up adds nothing to it.
• E. $\frac{10}{3}\text{m}$ — Height Used
  The same division by $30$ applied to the correct $10\ \text{m}$. Vertical height plays no part in the toppling moment.

Common Mistake (⚠️):
Trying to involve the vertical distance $30\text{m}$ or crane height $100\text{m}$ in the moment calculation.

Takeaway (📌):
The moment of a vertical force (weight) depends strictly on its horizontal perpendicular distance from the pivot.

Correct Answer: A. 1 mm

Fastest Approach (🚀):
$4(f L/1000) = 0.08 \implies f L = 20$. For $f > 20,000\text{Hz}$, $L = 20/f < 20/20,000 = 0.001\text{m} = 1\ \text{mm}$. Matches Option A.

Step-by-Step Breakdown:

1. Setting up the Lift Equation

The lift from one wing is $F = \frac{f L}{1000}$.
The bee has 4 wings, so the total lift is:
$F_{\text{total}} = 4 \times \frac{f L}{1000} = \frac{f L}{250}$

2. Equating Lift and Weight

To fly, the total lift must balance the bee's weight ($0.08\text{N}$):
$\frac{f L}{250} = 0.08$
$f L = 0.08 \times 250 = 20$

3. Applying Human Hearing Range

The frequency must be so high that a human cannot hear it. The human hearing range is approximately $20\text{Hz}$ to $20,000\text{Hz}$.
Therefore, the bee's wingbeat frequency must be $f > 20,000\text{Hz}$.

4. Calculating Wing Length Bound

Since $L = \frac{20}{f}$ and $f > 20,000$:
$L < \frac{20}{20,000} = \frac{1}{1000}\text{m} = 1\ \text{mm}$

To be inaudible, the wing must be shorter than $1\ \text{mm}$.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. 4 mm — Missing Factor
  Forgetting that the bee has four wings. Using one wing gives $fL = 0.08 \times 1000 = 80$ and $L = 80/20\,000 = 4\text{ mm}$; with all four, $fL = 20$ and $L = 1\text{ mm}$.
• C. 10 mm — Wrong Frequency
  $10\text{ mm}$ corresponds to $f = 2000\text{ Hz}$. The top of human hearing is $20\,000\text{ Hz}$, which is the frequency the question is asking about.
• D. 0.01 mm — Unit Conversion
  A factor of $100$ too small - a metre-to-millimetre slip. $L = 20/20\,000 = 0.001\text{ m}$, which is $1\text{ mm}$.
• E. 0.1 mm — Unit Conversion
  A factor of $10$ too small. $L = 0.001\text{ m} = 1\text{ mm}$.

Common Mistake (⚠️):
Forgetting that the bee has 4 wings (resulting in $4\ \text{mm}$) or using the lower bound of human hearing ($20\text{Hz}$) instead of the upper bound.

Takeaway (📌):
Always account for total force contributors (4 wings) and know basic physiological constants like the human hearing range ($20\text{Hz}$ to $20\text{kHz}$).

Correct Answer: A. 900 cm

Fastest Approach (🚀):
$1/R_{eq} = 10 \implies 1 + x/100 = 10 \implies x/100 = 9 \implies x = 900\ \text{cm}$. Matches Option A.

Step-by-Step Breakdown:

1. Modeling the Circuit

The original wire has a resistance of $R_w = 1\ \text{\Omega}$.
Each centimetre of insulation acts as a parallel resistor of $R_i = 100\ \text{\Omega}$.
If there are $x$ centimetres of insulation, there are $x$ such $100\ \text{\Omega}$ resistors in parallel with each other, and in parallel with the main wire.

2. Setting Up Equivalent Resistance

The total resistance of $x$ identical $100\ \text{\Omega}$ resistors in parallel is $100/x \Omega$.
The total equivalent resistance $R_{\text{eq}}$ of the entire system is:
$\frac{1}{R_{\text{eq}}} = \frac{1}{R_w} + \frac{x}{R_i}$
$\frac{1}{R_{\text{eq}}} = \frac{1}{1} + \frac{x}{100}$

3. Solving for $x$

We are given that the final resistance drops to $0.1\ \text{\Omega}$, so $\frac{1}{R_{\text{eq}}} = \frac{1}{0.1} = 10$.
$10 = 1 + \frac{x}{100}$
$9 = \frac{x}{100}$
$x = 900$

So $900\ \text{cm}$ of insulation are required.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. 9000 cm — Order of Magnitude
  A factor of ten out. $1/0.1 = 10$, so $1 + x/100 = 10$, giving $x = 900$.
• C. 100 cm — Incomplete Calculation
  $100\text{ cm}$ puts $100\ \Omega$ of insulation in parallel with the $1\ \Omega$ wire: $1/R = 1 + 1 = 2$, so $R = 0.5\ \Omega$, not $0.1$.
• D. 1000 cm — Overshoot
  $1000\text{ cm}$ gives $1/R = 1 + 10 = 11$, so $R \approx 0.091\ \Omega$ - past the target. The exact figure is $900\text{ cm}$.
• E. The resistance cannot drop that far. — Premature Rejection
  It can. Every parallel path lowers the total resistance, and there is no floor: $1/R = 1 + x/100$ can be made as large as you like.

Common Mistake (⚠️):
Adding the insulation resistance in series instead of parallel, or forgetting to include the original wire's $1\ \text{\Omega}$ in the parallel sum.

Takeaway (📌):
When adding components in parallel, conductances ($1/R$) simply add up: $G_{\text{total}} = G_1 + G_2 + \dots + G_N$.

Correct Answer: A. X moves, Y moves

Fastest Approach (🚀):
$F_{\text{applied}} = 25\text{N} > F_{\text{max}} = 5\text{N} \implies$ both move.
Matches Option A.

Step-by-Step Breakdown:

1. Calculating Maximum Static Friction

For each cube of weight $W = 10\text{N}$ and $\mu = 0.5$:
$F_{\text{max}} = \mu W = 0.5 \times 10\text{N} = 5\text{N}$

2. Comparing with Applied Force

The applied pulling force is $25\text{N}$.
Since $25\text{N} > 5\text{N}$, static friction is overcome for both cubes X and Y, causing both to move.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. X moves, Y remains stationary — Area Dependence
  Friction does not depend on contact area. $F_\text{max} = \mu N$ uses the weight, which is $10\text{ N}$ for both cubes, so both have the same $5\text{ N}$ limit and both move.
• C. X remains stationary, Y moves — Area Dependence
  The same area misconception the other way round. A larger base does not increase friction: $F_\text{max} = 0.5 \times 10 = 5\text{ N}$ for both.
• D. Both remain stationary — Comparison Error
  The applied force is $25\text{ N}$, five times the $5\text{ N}$ maximum friction each cube can muster, so neither stays put.
• E. There is not enough information. — Premature Rejection
  Everything needed is given: $F_\text{max} = \mu N = 0.5 \times 10 = 5\text{ N}$, and $25\text{ N}$ exceeds it. The base areas are not needed at all.

Common Mistake (⚠️):
Thinking pressure ($N/m^2$) alters total friction force (friction depends on $N$, not contact area).

Takeaway (📌):
Friction force $F = \mu N$ is independent of surface contact area.

Correct Answer: E. The ball does not reach the finish

Fastest Approach (🚀):
$KE_0 = 0.5\text{J} < mgh_{\text{peak}} = 10\text{J} \implies$ ball does not reach finish.
Matches Option E.

Step-by-Step Breakdown:

1. Initial Mechanical Energy

Set reference height $h = 0$ at starting point:
$E_{\text{init}} = \frac{1}{2} m v^2 + 0 = \frac{1}{2} (1\ \text{kg}) (1\text{ms}^{-1})^2 = 0.5\text{J}$

2. Energy Required to Cross Peak

The peak is $1\text{m}$ above starting point:
$PE_{\text{peak}} = m g h = 1\ \text{kg} \times 10\text{ms}^{-2} \times 1\text{m} = 10\text{J}$

3. Energy Barrier Comparison

Since $E_{\text{init}} = 0.5\text{J} < 10\text{J}$, the ball lacks sufficient kinetic energy to clear the peak energy barrier and rolls back before reaching the finishing point.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $40\text{ms}^{-1}$ — Squared Not Rooted
  Even ignoring the peak, $u^{2}+2gh=1+40=41$ and its square root is about $6.4$. This is the value under the root, not the speed.
• B. $20\text{ms}^{-1}$ — Squared Not Rooted
  Also the value under a root rather than a speed, and it ignores the $1\ \text{m}$ peak standing between start and finish.
• C. $\sqrt{40}\text{ms}^{-1}$ — Peak Ignored
  The right energy calculation from start to finish - but it assumes the ball gets there. Clearing the peak needs $mgh=10\ \text{J}$ and the ball starts with $\tfrac12(1)(1^{2})=0.5\ \text{J}$.
• D. $\sqrt{20}\text{ms}^{-1}$ — Wrong Height
  Measures the drop to the peak rather than to the finish, and again treats the peak as passable.

Common Mistake (⚠️):
Applying $v = \sqrt{u^2 + 2gh}$ between start and finish while ignoring the intermediate peak.

Takeaway (📌):
Always check energy barriers (intermediate peaks) before calculating final speed from potential drop.

Correct Answer: B. $2\text{ms}^{-2}$

Fastest Approach (🚀):
$a = \frac{50 - 10 - 20}{1 + 4 + 5} = \frac{20}{10} = 2\text{ms}^{-2}$.
Matches Option B.

Step-by-Step Breakdown:

1. Calculating Net Accelerating Force

• Weight of hanging mass $Z$: $W_Z = 50\text{N}$ (pulls right)
• Weight of hanging mass $X$: $W_X = 10\text{N}$ (opposes motion)
• Friction force on mass $Y$: $F_f = \mu W_Y = 0.5 \times 40\text{N} = 20\text{N}$ (opposes motion)

Net force driving the system:
$F_{\text{net}} = W_Z - W_X - F_f = 50 - 10 - 20 = 20\text{N}$

2. Total System Mass and Acceleration

Total weight $W_{\text{total}} = 10 + 40 + 50 = 100\text{N} \implies M_{\text{total}} = \frac{100}{10} = 10\ \text{kg}$.
$a = \frac{F_{\text{net}}}{M_{\text{total}}} = \frac{20\text{N}}{10\ \text{kg}} = 2\text{ms}^{-2}$

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $1\text{ms}^{-2}$ — Factor of Two
  Half the answer: the net force is $50-10-0.5(40)=20\ \text{N}$ and the accelerated mass is $\tfrac{10+40+50}{10}=10\ \text{kg}$.
• C. $3\text{ms}^{-2}$ — Friction Understated
  Understates the friction. On a rough plane with $\mu=0.5$ under a $40\ \text{N}$ weight, friction is $0.5\times40=20\ \text{N}$, not $10\ \text{N}$.
• D. $4\text{ms}^{-2}$ — Friction Omitted
  Omits friction entirely: $\tfrac{50-10}{10}=4$. Y rests on a rough plane, so $20\ \text{N}$ opposes the motion.
• E. None of the above — Premature Rejection
  $2\ \text{ms}^{-2}$ is on the list, so nothing forces this option.

Common Mistake (⚠️):
Forgetting to include mass $Y$ in the total accelerated mass $M_{\text{total}}$.

Takeaway (📌):
Acceleration of connected strings $a = \frac{F_{\text{net}}}{\sum m_i}$.

Correct Answer: B. $\frac{F}{8}$ pulling up

Fastest Approach (🚀):
Compound ratio $= \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} F$ up.
Matches Option B.

Step-by-Step Breakdown:

1. Analysing First Lever

Lever 1 pivots with arm lengths $1$ (where force $F$ is applied, pulling up) and $4$ (where the connecting string attaches). Taking moments about the pivot: $F \times 1 = T \times 4$, so the string tension is $T = \frac{F}{4}$. Since $F$ pulls up on the short arm, the long arm — and the string hanging from it — is pulled down.

2. Analysing Second Lever

Lever 2 pivots with arm lengths $1$ (where the string from Lever 1 pulls down) and $2$ (where output point X is located). Taking moments about this pivot: $T \times 1 = F_X \times 2$, so:
$F_X = \frac{T}{2} = \frac{F/4}{2} = \frac{F}{8}$
Since the string pulls down on the short arm, the long arm at X swings upward.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\frac{F}{8}$ pulling down — Direction Reversed
  Right magnitude, wrong direction. The second lever reverses the sense again, so the string pulls X upward.
• C. $\frac{8}{F}$ pulling up — Units Wrong
  Dimensionally impossible: dividing a pure number by a force does not give a force. A mechanical advantage multiplies $F$, it does not divide into it.
• D. $\frac{8}{F}$ pulling down — Units Wrong
  The same dimensional fault, with the direction wrong as well.
• E. None of the above — Premature Rejection
  $\tfrac{F}{8}$ pulling up is listed. The two lever ratios compound as $\tfrac14\times\tfrac12=\tfrac18$.

Common Mistake (⚠️):
Multiplying mechanical advantages ($4 \times 2 = 8$) into numerator instead of denominator.

Takeaway (📌):
Compound lever mechanical advantage is the product of individual lever ratios.

Correct Answer: A. 13%

Fastest Approach (🚀):
$Q = 300 \times 5 = 1500\text{J} \implies \frac{1500}{11500} \approx 13\%$.
Matches Option A.

Step-by-Step Breakdown:

1. Calculating Temperature Rise

Plastic deformation angle $\theta = 90^\circ = \frac{\pi}{2}\ \text{rad}$.
Given rate: $10\text{K}$ per $\pi\text{rad}$.
$\Delta T = 10 \times \left(\frac{\pi/2}{\pi}\right) = 5\text{K}$

2. Calculating Thermal Energy Produced

Using heat capacity $C = 300\text{JK}^{-1}$:
$Q = C \Delta T = 300\text{JK}^{-1} \times 5\text{K} = 1500\text{J}$

3. Finding Percentage of Kinetic Energy

$\text{Percentage} = \frac{1500\text{J}}{11,500\text{J}} \times 100\% \approx 13.04\% \approx 13\%$

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. 15% — Arithmetic Error
  $15\%$ of $11\,500\text{ J}$ is $1725\text{ J}$. The axle absorbs $Q = 300 \times \Delta T$ with $\Delta T = 10 \times \tfrac{\pi/2}{\pi} = 5\text{ K}$, so $Q = 1500\text{ J}$ and the share is $13\%$.
• C. 17% — Arithmetic Error
  $17\%$ would need $1955\text{ J}$. The $90^\circ$ twist is $\tfrac{\pi}{2}$ radians, which is half of the $\pi$ radians that give $10\text{ K}$, so $\Delta T = 5\text{ K}$.
• D. 19% — Angle Conversion
  $19\%$ would need $2185\text{ J}$, close to the $2 \times 1500\text{ J}$ you get by treating $90^\circ$ as $\pi$ radians rather than $\tfrac{\pi}{2}$.
• E. None of the above — Premature Rejection
  $13\%$ is available: $1500/11\,500 = 13.04\%$.

Common Mistake (⚠️):
Forgetting to convert $90^\circ$ to radians.

Takeaway (📌):
Percentage conversion $= \frac{Q_{\text{thermal}}}{E_{\text{kinetic}}} \times 100\%$.

Correct Answer: B. Yes, and my finger does not need to move downwards at any point, but I must move very slowly

Fastest Approach (🚀):
Quasi-static movement $\implies$ finger moves upwards very slowly.
Matches Option B.

Step-by-Step Breakdown:

1. Dynamics of Spring-Mass Motion

If the finger is moved abruptly upward, the spring contracts, releasing stored elastic energy into kinetic energy, causing the mass to oscillate vertically with downward velocity phases.

2. Quasi-Static Condition

To move the mass upward by $10\ \text{cm}$ without ever acquiring downward velocity, the motion must be performed quasi-statically (infinitely slowly).
This ensures zero kinetic energy buildup and continuous force equilibrium at all points, with the finger moving upwards continuously.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. Yes, and my finger does not need to move downwards at any point — Ignores Oscillation
  Moving at any finite speed leaves the spring out of equilibrium and sets it oscillating, and in an oscillation the weight does travel downwards. Only a quasi-static lift avoids that.
• C. Yes, but my finger must move downwards at some point — Unnecessary Motion
  The finger never needs to descend. The weight is unchanged, so the spring's extension at equilibrium is unchanged, and raising the finger $10\text{ cm}$ slowly raises the weight $10\text{ cm}$.
• D. Yes, but my finger must move downwards at some point, but I must move very slowly — Unnecessary Motion
  The slow motion is the right instinct, but the downwards move is not needed: the equilibrium extension depends only on the weight, which never changes.
• E. No — Premature Rejection
  It is possible - raise the finger slowly enough that the spring stays in equilibrium at every instant, and the weight rises monotonically.

Common Mistake (⚠️):
Thinking the finger must move downwards first to relieve tension.

Takeaway (📌):
Eliminating oscillations requires quasi-static (very slow) movement.

Correct Answer: F. $V_X = -2\text{ms}^{-1}, V_Y = 1.75\text{ms}^{-1}$

Fastest Approach (🚀):
Check Option F: $P = -2 + 7 = 5$, $KE = 2 + 6.125 = 8.125 = 0.65 KE_0$.
Matches Option F.

Step-by-Step Breakdown:

1. Initial System Values

• Initial Momentum: $P_{\text{init}} = 1\ \text{kg} \times 5\text{ms}^{-1} = 5\ \text{kg ms}^{-1}$
• Initial Kinetic Energy: $KE_{\text{init}} = \frac{1}{2} (1) (5^2) = 12.5\text{J}$
• Target Final Energy ($35\%$ loss $\implies 65\%$ retained):

$KE_{\text{final}} = 0.65 \times 12.5 = 8.125\text{J}$

2. Testing Option F ($V_X = -2\text{ms}^{-1}, V_Y = 1.75\text{ms}^{-1}$)

1. Momentum: $P_{\text{final}} = 1(-2) + 4(1.75) = -2 + 7 = 5\ \text{kg ms}^{-1}$ (Conserved)
2. Energy: $KE_{\text{final}} = \frac{1}{2}(1)(-2)^2 + \frac{1}{2}(4)(1.75)^2 = 2 + 2(3.0625) = 8.125\text{J}$ (Conserved with $35\%$ loss)

Matches Option F.

Why the Other Options Are Wrong (❌):

• A. $V_X = 1\text{ms}^{-1}, V_Y = 2\text{ms}^{-1}$ — Momentum Not Conserved
  Momentum is not conserved: $1(1) + 4(2) = 9\text{ kg m s}^{-1}$, but the initial total is $1(5) = 5\text{ kg m s}^{-1}$. Check momentum before energy - it rules most options out immediately.
• B. $V_X = 1.8\text{ms}^{-1}, V_Y = 1.8\text{ms}^{-1}$ — Momentum Not Conserved
  $1(1.8) + 4(1.8) = 9\text{ kg m s}^{-1}$, against an initial $5$. Equal final speeds would also mean the spheres move together, which is a perfectly inelastic collision, not a $35\%$ loss.
• C. $V_X = -2\text{ms}^{-1}, V_Y = 2\text{ms}^{-1}$ — Momentum Not Conserved
  $1(-2) + 4(2) = 6\text{ kg m s}^{-1}$, not $5$. The energy is also wrong: this gives $10\text{ J}$ against the required $8.125\text{ J}$.
• D. $V_X = -1\text{ms}^{-1}, V_Y = 3\text{ms}^{-1}$ — Momentum Not Conserved
  $1(-1) + 4(3) = 11\text{ kg m s}^{-1}$, more than twice the initial $5$, and the kinetic energy rises to $18.5\text{ J}$ - a collision cannot create energy.
• E. $V_X = 0\text{ms}^{-1}, V_Y = 3\text{ms}^{-1}$ — Momentum Not Conserved
  $0 + 4(3) = 12\text{ kg m s}^{-1}$, against an initial $5$, and $18\text{ J}$ of kinetic energy against an initial $12.5\text{ J}$.

Common Mistake (⚠️):
Testing energy without checking momentum conservation.

Takeaway (📌):
In inelastic collisions, momentum MUST be conserved perfectly ($P_{\text{final}} = P_{\text{init}}$).

Correct Answer: A. 1

Fastest Approach (🚀):
Arrangement 1 allows continuous beam bending downwards.
Matches Option A.

Step-by-Step Breakdown:

1. Why Arrangement 1 Works

To bend a laser beam downward without any internal reflections, light must enter glass blocks at non-zero incidence angles, refracting toward the normal on entry ($n > 1$) and away from the normal on exit. Arrangement 1 uses a sequence of small blocks, each tilted slightly more than the last, so every interface refracts the beam a little further — the cumulative effect of these small-angle refractions gradually turns the beam from horizontal to vertical.

2. Why Arrangements 2 and 3 Fail

Arrangement 2 is a single block with two parallel flat faces. For a parallel-sided slab, the refraction at entry is exactly cancelled by the refraction at exit (equal and opposite bending), so the beam re-emerges travelling in its original horizontal direction — only shifted sideways, never turned downward.
Arrangement 3 is a rectangular block whose internal faces meet only at right angles. A beam entering perpendicular to one face travels straight through undeviated at every internal boundary, so there is no angled surface available to redirect it; the beam simply continues horizontally rather than turning to exit downward.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. 2 — Parallel Faces
  A slab with parallel faces shifts a ray sideways but returns it to its original direction - the bend on entry is undone on exit. It cannot turn the beam through a right angle.
• C. 3 — Reflection Excluded
  The L-block's faces are flat and mutually perpendicular. A ray entering along the normal of one face is not bent at all, and it then meets the perpendicular face at grazing incidence. The right-angle turn a real prism makes uses reflection, which this question rules out.
• D. 1, 3 — One Wrong Included
  Arrangement 1 does work, but 3 does not: perpendicular flat faces cannot turn a beam by refraction alone.
• E. None of them — Premature Rejection
  Arrangement 1 does work. Each block in the fan is entered near normal incidence and turns the beam a little, and enough small refractions accumulate to a right angle.

Common Mistake (⚠️):
Assuming refraction cannot bend light by large angles without reflection.

Takeaway (📌):
Successive small-angle refractions can redirect light beams by arbitrary angles.

Correct Answer: F. $V_P = 0.6\text{ms}^{-1}, V_Q = 1.6\text{ms}^{-1}$

Fastest Approach (🚀):
$3 V_P + 2(V_P + 1) = 5 \implies 5 V_P = 3 \implies V_P = 0.6, V_Q = 1.6$.
Matches Option F.

Step-by-Step Breakdown:

1. Initial Momentum Calculation

Taking direction of P as positive ($+x$):
$P_{\text{init}} = m_P u_P + m_Q u_Q = 3(3) + 2(-2) = 9 - 4 = 5\ \text{kg ms}^{-1}$

2. Setting Up Simultaneous Equations

From momentum conservation:
$3 V_P + 2 V_Q = 5$

We are given that after collision, both move in the same direction with speed difference $1\text{ms}^{-1}$ ($V_Q = V_P + 1$):
$3 V_P + 2 (V_P + 1) = 5$
$5 V_P + 2 = 5 \implies 5 V_P = 3 \implies V_P = 0.6\text{ms}^{-1}$
$V_Q = 0.6 + 1 = 1.6\text{ms}^{-1}$

Matches Option F.

Why the Other Options Are Wrong (❌):

• A. $V_P = 0.1\text{ms}^{-1}, V_Q = 1.1\text{ms}^{-1}$ — Momentum Not Conserved
  Momentum after would be $3(0.1) + 2(1.1) = 2.5\text{ kg m s}^{-1}$, but momentum is conserved and the initial total is $3(3) - 2(2) = 5\text{ kg m s}^{-1}$.
• B. $V_P = 0.2\text{ms}^{-1}, V_Q = 1.2\text{ms}^{-1}$ — Momentum Not Conserved
  $3(0.2) + 2(1.2) = 3.0\text{ kg m s}^{-1}$, against the required $5$.
• C. $V_P = 0.3\text{ms}^{-1}, V_Q = 1.3\text{ms}^{-1}$ — Momentum Not Conserved
  $3(0.3) + 2(1.3) = 3.5\text{ kg m s}^{-1}$, against the required $5$.
• D. $V_P = 0.4\text{ms}^{-1}, V_Q = 1.4\text{ms}^{-1}$ — Momentum Not Conserved
  $3(0.4) + 2(1.4) = 4.0\text{ kg m s}^{-1}$, against the required $5$.
• E. $V_P = 0.5\text{ms}^{-1}, V_Q = 1.5\text{ms}^{-1}$ — Momentum Not Conserved
  $3(0.5) + 2(1.5) = 4.5\text{ kg m s}^{-1}$ - close, but momentum must come to exactly $5$, which it does only at $V_P = 0.6$.

Common Mistake (⚠️):
Miscalculating initial momentum sign for Q ($3(3) + 2(2) = 13$ instead of $5$).

Takeaway (📌):
Always assign explicit signs ($+x, -x$) to initial velocity vectors.

Correct Answer: C. $a = \frac{3g}{5}$

Fastest Approach (🚀):
$a = g(4/5 - 1/5) = \frac{3}{5}g$.
Matches Option C.

Step-by-Step Breakdown:

1. Trigonometry of Incline Angle

Given $\tan\theta = \frac{4}{3}$, constructing a $3-4-5$ right triangle yields:
$\sin\theta = \frac{4}{5}, \quad \cos\theta = \frac{3}{5}$

2. Force Balance Along Incline

• Parallel component of weight: $W_{\parallel} = m g \sin\theta$
• Normal force: $N = m g \cos\theta$
• Friction force: $F_f = \mu N = \mu m g \cos\theta$

Net force down slope:
$F_{\text{net}} = m a = m g \sin\theta - \mu m g \cos\theta$
$a = g (\sin\theta - \mu \cos\theta)$

3. Substituting Numerical Values

$a = g \left(\frac{4}{5} - \frac{1}{3} \times \frac{3}{5}\right) = g \left(\frac{4}{5} - \frac{1}{5}\right) = \frac{3}{5}g$

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $a = \frac{g}{5}$ — Coefficient Dropped
  The friction term has lost its coefficient. With $\mu = \tfrac13$ and $\cos\theta = \tfrac35$, friction contributes $\tfrac13 \times \tfrac35 = \tfrac15 g$, not $\tfrac35 g$.
• B. $a = \frac{2g}{5}$ — Arithmetic Error
  Would need the friction term to be $\tfrac25 g$. From $\tan\theta = \tfrac43$ we get $\sin\theta = \tfrac45$ and $\cos\theta = \tfrac35$, so $a = g(\tfrac45 - \tfrac13 \cdot \tfrac35) = \tfrac35 g$.
• D. $a = \frac{4g}{5}$ — Friction Ignored
  Friction ignored entirely: $a = g\sin\theta = \tfrac45 g$. The rough plane takes $\tfrac15 g$ back off that.
• E. $a = g$ — Component Ignored
  The full $g$ is the free-fall value. On a slope only the component along the plane acts, $g\sin\theta = \tfrac45 g$, and friction reduces it further.
• F. $a = \frac{6g}{5}$ — Physically Impossible
  Larger than $g\sin\theta = \tfrac45 g$, which is impossible: friction can only reduce the acceleration down the slope, never increase it.

Common Mistake (⚠️):
Forgetting mass cancels out of acceleration equation $a = g(\sin\theta - \mu\cos\theta)$.

Takeaway (📌):
Incline acceleration without external forces is independent of mass $m$.

Correct Answer: B. Decreasing A, increasing B

Fastest Approach (🚀):
Decrease A (lower series drop), Increase B (divert current to C).
Matches Option B.

Step-by-Step Breakdown:

1. Circuit Topology Analysis

• Resistor A is in series with the main supply.
• Resistor B and Bulb C are in parallel with each other.

2. Maximizing Voltage Across Bulb C

1. Decreasing A: Reduces total series resistance, decreasing voltage drop across A and increasing available voltage to the parallel network.
2. Increasing B: Increases parallel network resistance, forcing a larger fraction of current through Bulb C.

Combining both actions maximizes voltage and brightness of Bulb C.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. Increasing A, increasing B — Series Wrong Way
  Increasing the series resistor A drops more of the supply before the bulb, so C dims. Only the change to B is right.
• C. Decreasing A, Decreasing B — Parallel Wrong Way
  Decreasing the parallel resistor B diverts more current away from C - taken far enough it shorts the bulb out. Only the change to A is right.
• D. Increasing A, Decreasing B — Both Wrong Way
  Both changes work against the bulb: more voltage dropped in series and more current bypassing it in parallel.
• E. Increasing A, holding B constant — Series Wrong Way
  Increasing A alone can only dim C, and holding B fixed gives up the one gain that was available.

Common Mistake (⚠️):
Decreasing B, which shorts out Bulb C.

Takeaway (📌):
To maximize current in a parallel branch, increase resistance of parallel sub-branches.

Correct Answer: C. 1, 1, -1, 1

Fastest Approach (🚀):
Dimensional analysis matches Option C.

Matches Option C.

Step-by-Step Breakdown:

1. Base Dimensions of Variables

• Terminal velocity $[V_t] = \text{L T}^{-1}$
• Area $[A] = \text{L}^2$
• Weight $[W] = \text{M L T}^{-2}$
• Gravity $[g] = \text{L T}^{-2} = \text{N kg}^{-1}$
• Constant $[C] = \text{L}^{-1} \text{T}^{-1} \text{M}^{-1}$

2. Equating Exponents

$[L T^{-1}] = (L^2)^\alpha (M L T^{-2})^\beta (L T^{-2})^\gamma (L^{-1} T^{-1} M^{-1})^\delta$

Matching the exponents of each base dimension gives three simultaneous equations:

• Mass (M): $0 = \beta - \delta$
• Length (L): $1 = 2\alpha + \beta + \gamma - \delta$
• Time (T): $-1 = -2\beta - 2\gamma - \delta$

The M equation gives $\beta = \delta$. Substituting into the L and T equations and testing against the five options, only $\alpha=1, \beta=1, \gamma=-1, \delta=1$ satisfies all three simultaneously.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. -1, -1, 1, 1 — Dimensional Mismatch
  Dimensionally wrong on every count. With $\alpha = \beta = -1$ the metres come to $-3$, where $V_t$ needs $\text{m}^1$.
• B. 1, 1, -1, -1 — Dimensional Mismatch
  $\delta = -1$ leaves $\text{kg}^2$ in the result: $W$ contributes one kilogram and $C^{-1}$ another. $V_t$ must be free of mass.
• D. 1, 1, 1, -1 — Dimensional Mismatch
  $\gamma = +1$ gives $\text{s}^{-4}$ overall instead of $\text{s}^{-1}$, and $\delta = -1$ leaves $\text{kg}^2$ uncancelled.
• E. 1, -1, -1, -1 — Dimensional Mismatch
  The kilograms do cancel here, but the metres come to $3$ and the seconds to $+5$: $V_t$ needs $\text{m s}^{-1}$.

Common Mistake (⚠️):
Misidentifying dimensions of constant $C$.

Takeaway (📌):
Equating dimensions $[M^a L^b T^c]$ forms a system of linear equations in unknown exponents.