ESAT Paper 1 Physics Worked Solutions

Correct Answer: D. 5.00 radians/second

Fastest Approach (🚀):
Efficiency equation: $P_{\text{out}} = \eta \cdot P_{\text{in}} \implies 100 = 0.80 \cdot P_{\text{in}} \implies P_{\text{in}} = 125 \ \text{W}$.
Torque $\tau = F \cdot r = 100 \times 0.25 = 25 \text{N m}$.
Rotational power $P_{\text{in}} = \tau \omega \implies 125 = 25 \omega \implies \omega = 5.00\ \text{rad/s}$.

Matches Option D.

Step-by-Step Breakdown:

1. Rotational Mechanics & Efficiency Principles

A. Mechanical Efficiency Law

$\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \implies P_{\text{in}} = \frac{P_{\text{out}}}{\eta}$

B. Rotational Torque Formula

$\tau = F \cdot r$

C. Rotational Power Formula

$P_{\text{mechanical}} = \tau \cdot \omega \quad (\omega \text{in rad/s})$

2. System Parameters & Input Power Calculation

• Electrical output required: $P_{\text{out}} = 100 \ \text{W}$.
• Mechanical efficiency: $\eta = 80 \% = 0.80$.
• Mechanical input power required:

$P_{\text{in}} = \frac{100 \ \text{W}}{0.80} = 125 \ \text{W}$

3. Torque & Angular Speed Calculation

• Force applied: $F = 100 \ \text{N}$. Radius: $r = 25 \ \text{cm} = 0.25 \ \text{m}$.
• Torque generated: $\tau = 100 \times 0.25 = 25 \text{N m}$.
• Solve for angular velocity $\omega$:

$P_{\text{in}} = \tau \cdot \omega \implies 125 = 25 \cdot \omega \implies \omega = \frac{125}{25} = 5.00\ \text{rad/s}$

4. Option Matching

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. 0.51 radians/second — Incomplete Calculation
  Failing to account for mechanical efficiency.
• B. 77 radians/second — Misapplied Formula
  Multiplying by the efficiency instead of dividing to find input power.
• C. 36 radians/second — Unit Conversion Error
  Using the diameter instead of the radius for the torque calculation.
• E. 2 radians/second — Conceptual Misunderstanding
  Dividing by the efficiency twice.

Common Mistake (⚠️):
Forgetting to account for efficiency or applying it backward by multiplying $100 \ \text{W}$ by 0.8 instead of dividing.

Takeaway (📌):
Always apply efficiency to the input power side of the energy conservation equation.

Correct Answer: C. $7.5\times10^{19}$

Fastest Approach (🚀):
Original resistance $R_0 = 2.0\ \Omega$.
Doubling radius increases area by 4, so new resistance is $R = 2.0 / 4 = 0.5\ \Omega$.
Ohm's law: $I = V / R = 6 / 0.5 = 12 \ \text{A} = 12 \text{C/s}$.
Number of electrons: $N = \frac{12}{1.6 \times 10^{-19}} = 7.5 \times 10^{19}\ \text{electrons/sec}$.

Matches Option C.

Step-by-Step Breakdown:

1. Resistivity, Ohm's Law & Charge Quantization Principles

A. Pouillet's Resistivity Law

$R = \rho \frac{L}{A} = \frac{\rho L}{\pi r^2}$

B. Ohm's Law

$I = \frac{V}{R}$

C. Charge Quantization Law

$I = \frac{\Delta Q}{\Delta t} = N \cdot e \quad (e = 1.6 \times 10^{-19}C)$

2. Resistance Scaling with Wire Radius

• Original resistance $R_0 = 2.0\ \Omega$. Radius $r_0 = 1$ cm.
• Doubling radius $r_1 = 2r_0 \implies A_1 = \pi(2r_0)^2 = 4\pi r_0^2 = 4A_0$.
• New resistance:

$R_{\text{new}} = \frac{R_0}{4} = \frac{2.0}{4} = 0.5\ \Omega$

3. Current & Electron Flow Calculation

• Voltage $V = 6 \ \text{V}$.
• New current: $I = \frac{6}{0.5} = 12 \ \text{A} = 12 \text{C/s}$.
• Number of electrons per second $N$:

$N = \frac{I}{e} = \frac{12}{1.6 \times 10^{-19}} = 7.5 \times 10^{19}\ \text{electrons/second}$

4. Option Matching

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $1.9\times10^{19}$ — Unit Conversion Error
  Assuming resistance is inversely proportional to radius rather than area.
• B. $3.8\times10^{19}$ — Unit Conversion Error
  Forgetting to halve the resistance when the radius doubles.
• D. $1.5\times10^{20}$ — Unit Conversion Error
  Squaring the current instead of simply dividing by the charge.
• E. $4.7 \times 10^{20}$ — Unit Conversion Error
  Using the wrong charge constant or making a power of 10 order-of-magnitude error.

Common Mistake (⚠️):
Assuming resistance is inversely proportional to radius rather than the square of the radius.

Takeaway (📌):
When a wire's radius changes by a factor of $x$, its resistance changes by a factor of $1/\text{x^2}$.

Correct Answer: B. $\sqrt[4]{\Omega(1-a)/\sigma}$

Fastest Approach (🚀):
Absorbed radiation fraction: $(1 - a)$. Total absorbed power $= \Omega(1 - a)$.
Stefan-Boltzmann law: $P = \sigma T_s^4$.
Equate: $\sigma T_s^4 = \Omega(1 - a) \implies T_s = \sqrt[4]{\frac{\Omega(1-a)}{\sigma}}$.

Matches Option B.

Step-by-Step Breakdown:

1. Stefan-Boltzmann Law & Albedo Definitions

A. Albedo Factor ($a$)

The fraction of incident radiation reflected by a body. The absorbed fraction is $(1 - a)$.

B. Stefan-Boltzmann Law

Power emitted per unit surface area by a black body at temperature $T$ is $P = \sigma T^4$, where $\sigma$ is the Stefan-Boltzmann constant.

C. Radiative Equilibrium Principle

$\text{Absorbed Flux} = \text{Emitted Flux}$

2. Energy Conservation Balance

• Incoming solar radiation $= \Omega$.
• Reflected flux $= a \Omega \implies$ Absorbed flux $= \Omega(1 - a)$.
• Emitted thermal black-body flux $= \sigma T_s^4$.

3. Solving for Surface Temperature $T_s$

$\sigma T_s^4 = \Omega(1 - a)$
$T_s^4 = \frac{\Omega(1 - a)}{\sigma}$
$T_s = \sqrt[4]{\frac{\Omega(1 - a)}{\sigma}}$

4. Option Matching

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $\sqrt[4]{\Omega(a-1)/\sigma}$ — Sign Error
  Inverting the absorption fraction to (a-1), which incorrectly gives a negative value.
• C. $\sqrt[4]{\Omega(a-1)}$ — Misapplied Formula
  Forgetting to divide by the Stefan-Boltzmann constant in the formula.
• D. $\sqrt[4]{\Omega(1-a)}$ — Misapplied Formula
  Forgetting to divide by the Stefan-Boltzmann constant in the final rearranged formula.
• E. $\sqrt[4]{\Omega/\sigma}$ — Conceptual Misunderstanding
  Ignoring the albedo factor completely and assuming 100% absorption.

Common Mistake (⚠️):
Using the albedo factor $a$ as the absorbed fraction rather than the reflected fraction.

Takeaway (📌):
Albedo represents reflectivity, so always use $(1 - \text{albedo})$ for energy absorption.

Correct Answer: B. $\frac{12}{25}\pi \text{m s}^{-1}$

Fastest Approach (🚀):
Wave frequency $f = \frac{v}{\lambda} = \frac{2.4}{1.2} = 2 \ \text{Hz}$.
Angular frequency $\omega = 2\pi f = 4\pi\text{rad/s}$.
Amplitude $A = 120\ \text{mm} = 0.12 \ \text{m}$.
Maximum transverse speed $v_{\text{max}} = A\omega = 0.12 \times 4\pi = 0.48\pi = \frac{12}{25}\pi \text{m s}^{-1}$.

Matches Option B.

Step-by-Step Breakdown:

1. Wave Equation & SHM Velocity Definitions

A. Wave Relation

$v_{\text{wave}} = f \cdot \lambda \implies f = \frac{v_{\text{wave}}}{\lambda}$

B. Angular Frequency

$\omega = 2\pi f$

C. SHM Maximum Transverse Velocity

A string particle oscillating in transverse SHM $y(t) = A \sin(\omega t + \phi)$ has maximum speed:
$v_{\text{transverse max}} = A \omega$

---

2. Frequency & Angular Velocity Calculation

• Wave speed $v_{\text{wave}} = 2.4 \ \text{m/s}$, Wavelength $\lambda = 1.2 \ \text{m}$.
• Frequency $f = \frac{2.4}{1.2} = 2 \ \text{Hz}$.
• Angular frequency $\omega = 2\pi(2) = 4\pi\text{rad/s}$.

---

3. Transverse Speed Calculation

• Amplitude $A = 120\ \text{mm} = 0.12 \ \text{m}$.
• Maximum speed of point $Q$:

$v_{\text{max}} = A \omega = 0.12 \times 4\pi = 0.48\pi \text{ m/s} = \frac{12}{25}\pi \text{m s}^{-1}$

---

4. Option Matching

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $2.4 \text{ m s}^{-1}$ — Incomplete Calculation
  This is the wave's propagation speed, not the transverse particle speed.
• C. $\frac{3}{25}\pi \text{m s}^{-1}$ — Conceptual Misunderstanding
  A factor of 1/4 error in the final calculation.
• D. $\frac{6}{25}\pi \text{m s}^{-1}$ — Unit Conversion Error
  A factor of 1/2 error in the calculation, caused by missing the 2 in the angular frequency formula.
• E. $1.2 \text{m s}^{-1}$ — Conceptual Misunderstanding
  This is the wavelength, not a speed.

Common Mistake (⚠️):
Attempting to plug the $0.4 \ \text{m}$ distance into a wave equation to find a specific instantaneous velocity instead of recognizing it asks for the maximum possible speed.

Takeaway (📌):
For any particle in a transverse wave, the maximum speed is strictly dependent on amplitude and frequency, completely independent of its position along the wave.

Correct Answer: A. $12 e^{-0.45}\text{ V}$

Fastest Approach (🚀):
$V_0 = 12 \ \text{V}$. Time constant $\tau = RC = 5600 \times 0.02 = 112 \ \text{s}$.
$V(50) = 12 e^{-50/112} = 12 e^{-0.45}V$.

Matches Option A.

Step-by-Step Breakdown:

1. RC Circuit Theory & Exponential Decay Laws

A. Capacitor Discharge Equation

$V(t) = V_0 e^{-t / \tau}$

B. Time Constant Definition

$\tau = R C$

2. Time Constant Calculation

• Initial voltage $V_0 = 12 \ \text{V}$.
• Resistance $R = 5.6 \text{k}\Omega = 5600\ \Omega$.
• Capacitance $C = 20,000\ \mu F = 0.02 \ \text{F}$.
• Time constant $\tau = R C = 5600 \times 0.02 = 112 \ \text{s}$.

3. Exponent Evaluation & Voltage Substitution

Discharge time $t = 50 \ \text{s}$.
Exponent ratio: $\frac{t}{\tau} = \frac{50}{112} \approx 0.45$.
$V(50) = 12 e^{-50/112} \approx 12 e^{-0.45}V$

4. Option Matching

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $12 e^{0.45}\text{ V}$ — Unit Conversion Error
  Using an exponential growth formula instead of exponential decay.
• C. $6 e^{-0.45}\text{ V}$ — Conceptual Misunderstanding
  Halving the initial voltage incorrectly.
• D. $6 e^{0.45}\text{ V}$ — Unit Conversion Error
  Halving the initial voltage and using exponential growth.
• E. $24 e^{-0.45}\text{ V}$ — Conceptual Misunderstanding
  Doubling the initial voltage incorrectly.

Common Mistake (⚠️):
Confusing exponential decay $e^{-x}$ with exponential growth $e^x$.

Takeaway (📌):
Capacitor discharge always follows the decay formula $V(t) = V_0 e^{-t/\tau}$.

Correct Answer: D. It swings backwards and forwards hitting each plate in turn

Fastest Approach (🚀):
Trace the charge lifecycle. Contact with plate P1 transfers positive charge, causing repulsion from P1 to P2. Contact with earthed P2 neutralises it and induces negative charge, causing attraction back to P1. It swings backwards and forwards hitting each plate in turn.

Matches Option D.

Step-by-Step Breakdown:

1. Electrostatic Conduction & Induction Principles

A. Charge Transfer by Conduction

Contact between a conductor and a voltage source $V$ forces the conductor to acquire the same electrical potential and net charge sign as the source.

B. Coulomb Repulsion & Attraction

Like charges repel ($F = q E > 0$ away from high voltage plate), opposite charges attract.

C. Earthing / Grounding

Contact with an earthed conductor ($V = 0$) neutralises net positive charge and allows negative charge to be induced under an external field.

2. Phase-by-Phase Oscillation Mechanics

• Phase 1 (Initial Contact with $P_1$): Sphere touches $P_1$ at potential $+V$, acquiring positive charge $+q$.
• Phase 2 (Repulsion toward $P_2$): Like positive charges repel $\implies$ sphere experiences force $F = +q E$ away from $P_1$ toward earthed plate $P_2$.
• Phase 3 (Contact with $P_2$): Sphere strikes earthed plate $P_2$. $+q$ drains to ground. The strong uniform field from $P_1$ induces negative charge $-q'$ on the sphere.
• Phase 4 (Attraction back to $P_1$): Negative charge $-q'$ is attracted to positive plate $P_1$, swinging it back across.

3. Sustained Electrostatic Shuttle

The process repeats continuously, shuttling charge between plates $\implies$ sphere swings backwards and forwards hitting each plate in turn.

4. Option Matching

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. It remains stationary — Unit Conversion Error
  Ignoring the electrostatic forces transferred by physical contact.
• B. It swings towards $P_1$ and remains in contact with the plate — Conceptual Misunderstanding
  Assuming the sphere sticks due to attraction without realizing it acquires the same charge upon contact.
• C. It swings back towards $P_2$ and remains in contact with the plate — Unit Conversion Error
  Assuming the earthed plate permanently neutralises the sphere without inducing an opposite charge from the external field.
• E. None of the above — Conceptual Misunderstanding
  The sphere does indeed swing back and forth as described in D.

Common Mistake (⚠️):
Thinking the sphere will just stick to the earthed plate after losing its charge.

Takeaway (📌):
In an electric field, an earthed plate doesn't just neutralise an object. It provides charge to bring the object to $0 \ \text{V}$ locally, which in the presence of an external field leaves it with a net charge opposite to the high voltage plate.

Correct Answer: B. Charging the capacitor with a potential difference of $V/2$

Fastest Approach (🚀):
$E = \frac{1}{2} C V^2$.
Halving $V \implies E' = \frac{1}{4} E_0$ (75% decrease).
Halving $C \implies E' = \frac{1}{2} E_0$ (50% decrease).
Halving $V$ gives the greatest energy reduction.

Matches Option B.

Step-by-Step Breakdown:

1. Capacitor Energy Storage Laws

A. Capacitance Charge Relation

$Q = C V$

B. Energy Stored Formulas

$E = \frac{1}{2} Q V = \frac{1}{2} C V^2 = \frac{Q^2}{2C}$

2. Modification Analysis

Initial stored energy: $E_0 = \frac{1}{2} C V^2$.

• Option A ($V \to 2 \ \text{V}$): $E_A = \frac{1}{2} C (2 \ \text{V})^2 = 4 E_0$ (300% increase).
• Option B ($V \to V/2$): $E_B = \frac{1}{2} C \left(\frac{V}{2}\right)^2 = \frac{1}{4} E_0 = 0.25 E_0$ (75% decrease).
• Option C ($C \to 2C$): $E_C = \frac{1}{2} (2C) V^2 = 2 E_0$ (100% increase).
• Option D ($C \to C/2$): $E_D = \frac{1}{2} \left(\frac{C}{2}\right) V^2 = 0.5 E_0$ (50% decrease).

3. Comparison & Option Matching

Option B results in a 75% drop, which is the greatest energy reduction.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. Charging the capacitor with a potential difference of $2V$ — Conceptual Misunderstanding
  This modification increases the energy by 300%.
• C. Replacing the capacitor with one of capacitance $2C$ — Conceptual Misunderstanding
  This modification increases the energy by 100%.
• D. Replacing the capacitor with one of capacitance $C/2$ — Conceptual Misunderstanding
  This modification decreases the energy by 50%, which is less than the 75% decrease from halving V.
• E. None of the above — Conceptual Misunderstanding
  Option B is a valid answer as it results in a 75% reduction.

Common Mistake (⚠️):
Failing to realize that changing $V$ also changes $Q$, so $E = \frac{1}{2}CV^2$ must be used.

Takeaway (📌):
Energy stored in a capacitor depends quadratically on potential difference.

Correct Answer: D. $\frac{20}{3} \ \text{kg}$

Fastest Approach (🚀):
Assign direction signs (Right $+$, Left $-$):
Initial: $v_A = +4$, $v_B = -5$.
Final: $v_A' = -2$, $v_B' = +4$.
Masses: $m_A = 10$, $m_B = X$.

Conservation of momentum:
$10(4) + X(-5) = 10(-2) + X(4)$
$40 - 5X= -20 + 4X \implies 60 = 9X \implies X= 20/3 \ \text{kg}$

Matches Option D.

Step-by-Step Breakdown:

1. Conservation of Linear Momentum & Vector Direction Convention

A. Conservation Law

$p_{\text{initial total}} = p_{\text{final total}}$

B. 1D Directional Vector Mapping

Choose rightward direction as positive (+):

• Ball A initially moves right: $v_A = +4 \ \text{m/s}$.
• Ball B initially moves left: $v_B = -5 \ \text{m/s}$.
• Ball A reverses (moves left): $v_A' = -2 \ \text{m/s}$.
• Ball B reverses (moves right): $v_B' = +4 \ \text{m/s}$.

2. Momentum Balance Equation

Mass $m_A = 10 \ \text{kg}$, mass $m_B = X \ \text{kg}$.
$m_A v_A + m_B v_B = m_A v_A' + m_B v_B'$
$10(+4) + X(-5) = 10(-2) + X(+4)$
$40 - 5X= -20 + 4X$

3. Algebraic Solution for Mass $X$

$60 = 9X \implies X= \frac{60}{9} = \frac{20}{3} \ \text{kg}$

4. Option Matching

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $5 \ \text{kg}$ — Sign Error
  Making a sign error on the final velocity of A.
• B. $\frac{10}{3} \ \text{kg}$ — Sign Error
  Making a sign error on the initial velocity of B.
• C. $\frac{15}{2} \ \text{kg}$ — Sign Error
  Making a sign error on both reversed velocities.
• E. $25 \ \text{kg}$ — Sign Error
  Adding the momentums without respecting vector directions.

Common Mistake (⚠️):
Forgetting to make the initial velocity of B and the final velocity of A negative to account for the reversed directions.

Takeaway (📌):
Always draw a quick coordinate axis for 1D collision problems and explicitly assign positive and negative signs to velocity values before doing any algebra.

Correct Answer: A. $48 \ \text{J}$ and $12 \ \text{J}$

Fastest Approach (🚀):
$k = 30 \text{N cm}^{-1} = 3000 \text{N m}^{-1}$. $x = 0.2 \text{m}$.
Stored energy $E = \frac{1}{2}(3000)(0.2)^2 = 60 \ \text{J}$.
$m_B = 4 m_A \implies K_A = 4 K_B$.
$5 K_B = 60 \implies K_B = 12 \ \text{J}$ and $K_A = 48 \ \text{J}$.

Matches Option A.

Step-by-Step Breakdown:

1. Elastic Energy & Momentum-Kinetic Energy Relations

A. Hooke's Law Elastic Potential Energy

$E_{\text{elastic}} = \frac{1}{2} k x^2$

B. Kinetic Energy in Terms of Momentum

$K = \frac{p^2}{2m}$
When two bodies explode from rest, $p_A = p_B = p$, so kinetic energy is inversely proportional to mass:
$\frac{K_A}{K_B} = \frac{m_B}{m_A}$

2. Elastic Energy Calculation

• $k = 30 \text{N/cm} = 3000 \text{N/m}$.
• Compression $x = 20 \ \text{cm} = 0.20 \ \text{m}$.
• Total energy:

$E_{\text{elastic}} = \frac{1}{2} (3000) (0.20)^2 = 1500 \times 0.04 = 60 \ \text{J}$

3. Kinetic Energy Partitioning

• Mass ratio: $m_B = 4 m_A \implies K_A = 4 K_B$.
• $K_A + K_B = 60 \implies 5 K_B = 60 \implies K_B = 12 \ \text{J}$.
• $K_A = 48 \ \text{J}$.

4. Option Matching

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $12 \ \text{J}$ and $4 \ \text{J}$ — Unit Conversion Error
  Missing the 1/2 factor in the initial elastic energy calculation.
• C. $8 \ \text{J}$ and $2 \ \text{J}$ — Unit Conversion Error
  Making a unit conversion error with the spring constant (N/cm to N/m).
• D. $48 \ \text{J}$ and $36 \ \text{J}$ — Conceptual Misunderstanding
  Incorrectly assuming the energy is split proportionally to mass rather than inversely proportional.
• E. $12 \ \text{J}$ and $8 \ \text{J}$ — Incomplete Calculation
  An arbitrary arithmetic error during the energy partitioning.

Common Mistake (⚠️):
Assuming kinetic energy splits proportionally to mass instead of inversely proportional to mass.

Takeaway (📌):
In two-body momentum explosions, the lighter object carries the majority of kinetic energy.

Correct Answer: C. $\frac{40}{\sqrt{3}} \le u \le \frac{40+\sqrt{1600+152\sqrt{3}}}{2\sqrt{3}}$

Fastest Approach (🚀):
Time of flight $T = \frac{2 u \sin 30^\circ}{10} = \frac{u}{10}$.
Ball range $R = u \cos 30^\circ T = \frac{\sqrt{3} u^2}{20}$.
Truck back position $x_{\text{back}} = 20 T = 2u$. Truck front $x_{\text{front}} = 2u + 1.9$.
Land inside: $2u \le \frac{\sqrt{3} u^2}{20} \le 2u + 1.9$.
Left inequality: $2u \le \frac{\sqrt{3} u^2}{20} \implies u \ge \frac{40}{\sqrt{3}}$, which matches only Option C's lower bound.
Right inequality: $\frac{\sqrt{3} u^2}{20} \le 2u + 1.9 \implies \sqrt{3}u^2 - 40u - 38 \le 0 \implies u \le \frac{40+\sqrt{1600+152\sqrt{3}}}{2\sqrt{3}} \approx 24.0 \ \text{m/s}$.

Matches Option C.

Step-by-Step Breakdown:

1. 2D Projectile Kinematics & Moving Target Constraints

A. Vertical Motion Time of Flight

$T = \frac{2 u \sin\theta}{g}$

B. Horizontal Range Formula

$R = u \cos\theta \cdot T$

C. Landing Inequality

$x_{\text{back}}(T) \le R \le x_{\text{front}}(T)$

2. Time of Flight & Ball Range Formulas

• $\theta = 30^\circ$, $g = 10 \ \text{m/s}^2$.
• Time of flight $T = \frac{2 u \sin 30^\circ}{10} = \frac{u}{10}$ seconds.
• Ball horizontal range $R = u \left(\frac{\sqrt{3}}{2}\right) \left(\frac{u}{10}\right) = \frac{\sqrt{3} u^2}{20}$ metres.

3. Moving Truck Boundaries & Inequality System

• Truck speed $v = 20$ m/s, length $L = 1.9$ m.
• Truck back at time $T$: $x_{\text{back}} = 20 T = 2u$ metres.
• Truck front at time $T$: $x_{\text{front}} = 2u + 1.9$ metres.
• Landing inequality:

$2u \le \frac{\sqrt{3} u^2}{20} \le 2u + 1.9$

4. Lower Bound Derivation & Option Matching

Solve left inequality:
$2u \le \frac{\sqrt{3} u^2}{20} \implies 40 u \le \sqrt{3} u^2 \implies u \ge \frac{40}{\sqrt{3}}$

This lower bound, $\frac{40}{\sqrt{3}}$, matches only Option C among the five choices.

5. Upper Bound Derivation & Option Matching

Solve the right inequality:
$\frac{\sqrt{3} u^2}{20} \le 2u + 1.9 \implies \sqrt{3} u^2 - 40u - 38 \le 0$

By the quadratic formula, the relevant (upper) root is:
$u \le \frac{40 + \sqrt{1600 + 152\sqrt{3}}}{2\sqrt{3}} \approx 24.0 \ \text{m/s}$

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $\frac{20}{\sqrt{3}} \le u \le \frac{80}{\sqrt{3}} \sqrt{\frac{1600+152\sqrt{3}}{5\sqrt{3}}}$ — Unit Conversion Error
  Solving both sides of the inequality unnecessarily when the lower bound alone uniquely identifies the correct answer.
• B. $\frac{20}{\sqrt{3}} \le u \le \sqrt{\frac{1200+152\sqrt{3}}{2\sqrt{3}}}$ — Misapplied Formula
  An arithmetic error in the quadratic formula for the upper bound.
• D. $\frac{50}{\sqrt{3}} \le u \le \frac{40}{\sqrt{3}} \sqrt{\frac{1200+152\sqrt{3}}{2\sqrt{3}}}$ — Incomplete Calculation
  An arithmetic error in the lower bound derivation.
• E. $\frac{60}{\sqrt{3}} \le u \le \frac{40}{\sqrt{3}} \sqrt{\frac{1200+152\sqrt{3}}{2\sqrt{3}}}$ — Conceptual Misunderstanding
  Miscalculating the time of flight.

Common Mistake (⚠️):
Overcomplicating the relative velocity by trying to transform the entire problem into the truck's reference frame. It's often easier to keep everything grounded in the earth frame.

Takeaway (📌):
For moving target projectile problems, solve for the time of flight first, then evaluate the position boundaries independently.

Correct Answer: B. Features Normal Reaction Force, Frictional force (pointing left), Force from Spring (pointing left), Weight

Fastest Approach (🚀):
Carriage moves rightward.
Weight acts downwards, Normal reaction acts upwards.
Spring buffer pushes back to the left.
Sliding friction opposes motion and also acts to the left.
Both retarding forces point left.

Matches Option B.

Step-by-Step Breakdown:

1. Contact & Non-Contact Force Principles

A. Gravitational Force (Weight)

Acts vertically downwards through centre of mass ($W = mg$).

B. Normal Reaction Force ($N$)

Acts perpendicular to contact surface (vertically upwards).

C. Hooke's Law Spring Force

$F_s = -k x$. When compressed by rightward motion, buffer pushes back (leftwards).

D. Kinetic Friction ($f_k$)

Always opposes relative sliding velocity. For rightward motion, friction acts leftwards.

2. Directional Vector Assignment

• Carriage moves rightward ($+x$ direction).
• Vertical equilibrium: $N = mg$ ($+y$ and $-y$ forces balance).
• Horizontal forces:
• Spring force $F_s$: Points left ($-x$).
• Friction force $f_k$: Points left ($-x$).

3. Free Body Diagram Selection

Diagram B correctly shows $N$ (up), $W$ (down), $F_s$ (left), $f_k$ (left).

4. Option Matching

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. Features Normal Reaction Force, Force from Spring (pointing left), Frictional force (pointing right), Weight — Conceptual Misunderstanding
  Shows the frictional force pointing right (with the direction of motion) instead of left. Friction opposes relative sliding, so it should point the same way as the spring force, not against it.
• C. Features Weight (pointing up), Normal Reaction Force (pointing left), Frictional force (pointing left), Force from Spring (pointing down) — Sign Error
  Identifying the correct forces, but showing the spring force pointing in the wrong direction (down instead of left).
• D. Features Force from Spring (pointing up), Frictional force (pointing left), Normal Reaction Force (pointing left), Weight (pointing down) — Conceptual Misunderstanding
  Showing the spring force pointing up instead of left.
• E. Features Normal Reaction Force, Frictional force, Weight, Force from Spring — Sign Error
  Listing the correct forces but failing to specify their directional vectors.

Common Mistake (⚠️):
Assuming friction always acts in the opposite direction of the main applied force. Friction opposes relative motion, not other forces.

Takeaway (📌):
When an object is braking or decelerating, all contact retarding forces point backward.

Correct Answer: A. $\tau = -\mu \frac{dv}{dy}$ with $[\tau] = Pa$

Fastest Approach (🚀):
Stress is Force / Area $\implies Pa$. Formula is linear in $\mu$ and $\frac{dv}{dy}$.

Matches Option A.

Step-by-Step Breakdown:

1. Fluid Shear Stress & Viscosity Definitions

A. Shear Stress ($\tau$)

Tangential force per unit area parallel to fluid layers:
$[\tau] = \frac{\text{Force}}{\text{Area}} = \frac{N}{m^2} = Pa$

B. Newton's Law of Viscosity

In laminar fluid flow, shear stress is directly proportional to the velocity gradient:
$\tau = -\mu \frac{dv}{dy}$

2. Dimensional Verification

• $[\frac{dv}{dy}] = \frac{\text{m s}^{-1}}{m} = s^{-1}$.
• Dynamic viscosity $[\mu] = \text{Pa s} = \text{N s m}^{-2}$.
• Right Hand Side dimensions: $[\mu \frac{dv}{dy}] = (\text{Pa s}) \times (s^{-1}) = Pa$.
• Dimensions match Left Hand Side $[\tau] = Pa$.

3. Distractor Elimination

• Option B uses $\text{N m}^{-1}$ (surface tension, not stress).
• Options C & D use incorrect non-linear powers of viscosity $\mu^2$ or $\mu^{-1}$.

4. Option Matching

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $\tau = -\mu \frac{dv}{dy}$ with $[\tau] = \text{N m}^{-1}$ — Unit Conversion Error
  Using the units for surface tension (N m^-1) instead of shear stress (N m^-2).
• C. $\tau = -\mu^2 \frac{dv}{dy}$ with $[\tau] = Pa$ — Unit Conversion Error
  Using an incorrect non-linear power of viscosity (mu^2).
• D. $\tau = -\mu^{-1} \frac{dv}{dy}$ with $[\tau] = \text{N m}^{-1}$ — Unit Conversion Error
  Using an incorrect inverse power of viscosity.
• E. $\tau = -\mu^2 \frac{dv}{dy}$ with $[\tau] = \text{N m}^{-1}$ — Unit Conversion Error
  Using both the wrong units and the wrong power of viscosity.

Common Mistake (⚠️):
Confusing shear stress units ($\text{N m}^{-2}$) with surface tension units ($\text{N m}^{-1}$).

Takeaway (📌):
Stress of any kind is defined as Force per unit Area (Pascals).

Correct Answer: C. $170 \ \text{m}$

Fastest Approach (🚀):
$3t^2 = 60 + 2t + 2t^2 \implies t^2 - 2t - 60 = 0 \implies t = 1 + \sqrt{61} \approx 8.81 \ \text{s}$.
Distance from B's start $= 2(8.81) + 2(8.81)^2 \approx 173 \ \text{m} \approx 170 \ \text{m}$.

Matches Option C.

Step-by-Step Breakdown:

1. SUVAT Kinematics & Relative Displacement Equations

A. Position Formula

$x(t) = x_0 + v_0 t + \frac{1}{2} a t^2$

B. Intersection Condition

$x_A(t) = x_B(t)$

2. Position Equations Formulation

• Origin at Car A's starting point ($x_{0A} = 0$).
• Car A: $v_{0A} = 0, a_A = 6.0 \ \text{m/s}^2 \implies x_A(t) = 3t^2$.
• Car B: $x_{0B} = 60 \ \text{m}, v_{0B} = 2.0 \ \text{m/s}, a_B = 4.0 \ \text{m/s}^2 \implies x_B(t) = 60 + 2t + 2t^2$.

3. Collision Time & Car B Displacement

Set $x_A(t) = x_B(t)$:
$3t^2 = 60 + 2t + 2t^2 \implies t^2 - 2t - 60 = 0$
Solve quadratic:
$t = \frac{2 + \sqrt{4 - 4(-60)}}{2} = 1 + \sqrt{61} \approx 8.81 \ \text{s}$
Distance moved by Car B from ITS starting position:
$\Delta x_B = v_{0B} t + \frac{1}{2} a_B t^2 = 2(8.81) + 2(8.81)^2 \approx 172.85 \ \text{m} \approx 170 \ \text{m}$

4. Option Matching

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $460 \ \text{m}$ — Conceptual Misunderstanding
  Calculating the distance from Car A's origin instead of Car B's starting point, and making a calculation error.
• B. $230 \ \text{m}$ — Conceptual Misunderstanding
  Calculating the total distance from Car A's origin, not Car B's starting point.
• D. $1.7 \ \text{s}$ — Sign Error
  Making a sign error when solving the quadratic equation.
• E. $8.8 \ \text{s}$ — Unit Conversion Error
  This is the time to collision in seconds, not the distance.

Common Mistake (⚠️):
Calculating total distance from A's origin ($232.8 \ \text{m}$) instead of distance from B's starting point.

Takeaway (📌):
Pay close attention to the origin reference point when computing relative displacements.

Correct Answer: A. $2\pi x \sin(\theta)/\lambda$

Fastest Approach (🚀):
Path difference $\Delta L = x \sin\theta$.
Multiply by wave number $k = \frac{2\pi}{\lambda}$:
$\text{Phase Difference} = \frac{2\pi x \sin\theta}{\lambda}$

Matches Option A.

Step-by-Step Breakdown:

1. Geometric Path Difference ($\Delta L$)

When two parallel waves pass through slits separated by a distance $x$ at an angle $\theta$ to the normal, one wave must travel slightly further than the other to reach a distant screen.
If we drop a perpendicular line from the top slit to the path of the bottom wave, we form a small right-angled triangle.
The hypotenuse of this triangle is the slit separation, $x$.
The opposite side of this triangle represents the extra distance travelled by the lower wave, known as the path difference $\Delta L$.
Using basic trigonometry (SOH CAH TOA):
$\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{\Delta L}{x}$
Rearranging for $\Delta L$:
$\Delta L = x \sin\theta$

2. Converting Path to Phase Difference ($\Delta \phi$)

A full wavelength $\lambda$ corresponds to a full phase cycle of $2\pi$ radians.
Therefore, to convert any physical path difference into a phase angle, we multiply the path difference by the fraction of a full cycle it represents per metre (the wave number, $k = \frac{2\pi}{\lambda}$).
$\Delta \phi = \frac{2\pi}{\lambda} \cdot \Delta L$

3. Final Derivation

Substitute the geometric path difference $\Delta L = x \sin\theta$ into the phase equation:
$\Delta \phi = \frac{2\pi}{\lambda} (x \sin\theta) = \frac{2\pi x \sin\theta}{\lambda}$

4. Option Matching

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $\pi x \sin(\theta)/\lambda$ — Unit Conversion Error
  Using pi instead of 2*pi to represent a full phase cycle.
• C. $\pi x \sin(\theta)/2\lambda$ — Conceptual Misunderstanding
  Introducing a mistaken factor of 2 in the denominator.
• D. $2x \sin(\theta)/\pi\lambda$ — Unit Conversion Error
  Inverting the ratio, placing pi and lambda incorrectly.
• E. $x \sin(\theta)/\lambda$ — Conceptual Misunderstanding
  Forgetting the 2*pi factor entirely when converting from wavelengths to radians.

Common Mistake (⚠️):
Using just $\pi$ instead of $2\pi$ for a full phase cycle, or confusing the path difference formula with diffraction grating formulas.

Takeaway (📌):
To convert a physical path difference in metres to a phase difference in radians, simply multiply by $2\pi / \lambda$.

Correct Answer: B. $7\alpha+4\beta^-$

Fastest Approach (🚀):
Mass change: $235 - 207 = 28$. Alpha count $= 28 / 4 = 7$.
Seven alphas reduce atomic number by $7 \times 2 = 14 \implies 92 - 14 = 78$.
Final atomic number is $82 \implies 82 - 78 = 4\ \beta^-$ particles.
Total: $7\alpha + 4\beta^-$.

Matches Option B.

Step-by-Step Breakdown:

1. Conservation Laws in Radioactive Decay

A. Mass Number Conservation

$\sum A_{\text{reactants}} = \sum A_{\text{products}}$

B. Atomic Number Conservation

$\sum Z_{\text{reactants}} = \sum Z_{\text{products}}$

C. Decay Particle Signatures

• Alpha particle: $\alpha = {}^4_2 \text{He}$ ($A=4, Z=2$).
• Beta minus particle: $\beta^- = {}^{\ 0}_{-1}\text{e}$ ($A=0, Z=-1$).

2. Balancing Mass Numbers ($A$) for Alpha Count

Change in mass: $\Delta A = 235 - 207 = 28$.
Since $\beta^-$ has zero mass number:
$N_\alpha = \frac{\Delta A}{4} = \frac{28}{4} = 7\ \alpha \text{particles}$

3. Balancing Atomic Numbers ($Z$) for Beta Count

• Initial $Z_i = 92$.
• Reduction from 7 alpha decays: $7 \times 2 = 14 \implies Z_{\text{intermediate}} = 92 - 14 = 78$.
• Target $Z_f = 82$.
• Deficit in atomic number: $\Delta Z = 82 - 78 = +4$.
• Since each $\beta^-$ increases $Z$ by $+1$:

$N_\beta = 4\ \beta^-\text{particles}$
Total products: $7\alpha + 4\beta^-$.

4. Option Matching

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $8\alpha+6\beta^-$ — Conceptual Misunderstanding
  Miscalculating the initial mass difference.
• C. $6\alpha+4\beta^-$ — Unit Conversion Error
  A simple arithmetic error when subtracting the alpha particles' mass.
• D. $5\alpha+3\beta^-$ — Conceptual Misunderstanding
  Miscalculating both the mass difference and the required atomic number adjustment.
• E. $4\alpha+2\beta^-$ — Unit Conversion Error
  Misinterpreting the beta decay effect on the atomic number.

Common Mistake (⚠️):
Confusing Beta minus decay with Beta plus decay, causing the atomic number to be subtracted instead of added.

Takeaway (📌):
Always balance mass numbers first using alpha particles, as beta particles have a mass number of zero.

Correct Answer: A. 1.2 minutes

Fastest Approach (🚀):
As $t \to \infty$, the mass of U-234 reaches $16.0 \ \text{mg} \implies$ initial Pa-234 mass was $16.0 \ \text{mg}$.
Half-life occurs when half has decayed $\implies M_U = 8.0 \ \text{mg}$.
From table, $M_U = 8.0 \ \text{mg}$ at $t = 1.2\ \text{minutes}$.

Matches Option A.

Step-by-Step Breakdown:

1. Parent-Daughter Decay Kinematics

A. Mass Conservation Law

$M_{\text{parent}}(t) + M_{\text{daughter}}(t) = M_0$

B. Half-Life Definition

The time $t_{1/2}$ at which $M_{\text{parent}}(t_{1/2}) = \frac{1}{2} M_0 \implies M_{\text{daughter}}(t_{1/2}) = \frac{1}{2} M_0$.

2. Initial Mass Determination

• Decay process: $^{234}Pa \to {}^{234}\text{U} + \beta^-$.
• As $t \to \infty$, all parent nuclei have decayed, so $M_{\text{daughter}}(\infty) = M_0 = 16.0 \ \text{mg}$.

3. Half-Life Lookup

• Half-life target: $M_{\text{daughter}}(t_{1/2}) = \frac{16.0}{2} = 8.0 \ \text{mg}$.
• From the experiment data table, $M_{\text{daughter}} = 8.0 \ \text{mg}$ at $t = 1.2\ \text{minutes}$.
• Half-life $t_{1/2} = 1.2\ \text{minutes}$.

4. Option Matching

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. 2.4 minutes — Unit Conversion Error
  Misinterpreting the asymptote and taking half of an incorrect maximum value.
• C. 6.0 minutes — Conceptual Misunderstanding
  Reading the time for 15.5 mg instead of the halfway point.
• D. 9.6 minutes — Unit Conversion Error
  Misinterpreting the table data completely.
• E. 10.8 minutes — Unit Conversion Error
  Misinterpreting the table data.
• F. 12.0 minutes — Conceptual Misunderstanding
  Taking the final time recorded in the table as the half-life.

Common Mistake (⚠️):
Misinterpreting the table as the mass of the parent isotope remaining, which would lead to incorrect half-life calculations.

Takeaway (📌):
For a parent-daughter isotope pair, the sum of their masses (or number of nuclei) remains constant. Identify the asymptote to find the initial sample mass.

Correct Answer: A. Amplitude: $3 \ \text{m}$, Frequency: $\frac{1}{12 \times 3600}Hz$

Fastest Approach (🚀):
Amplitude $= \frac{16 - 10}{2} = 3 \ \text{m}$.
Period $T = 12 \times 3600 \ \text{s} \implies f = \frac{1}{12 \times 3600}Hz$.

Matches Option A.

Step-by-Step Breakdown:

1. Oscillatory Wave Parameter Laws

A. Equilibrium Midpoint

$y_0 = \frac{y_{\text{max}} + y_{\text{min}}}{2}$

B. Amplitude Formula

$A = y_{\text{max}} - y_0 = \frac{y_{\text{max}} - y_{\text{min}}}{2}$

C. Frequency Period Relation

$f = \frac{1}{T} \quad (T \text{in seconds})$

2. Amplitude Calculation

• $y_{\text{max}} = 16 \ \text{m}, y_{\text{min}} = 10 \ \text{m}$.
• $A = \frac{16 - 10}{2} = 3 \ \text{m}$.

3. Frequency Calculation

• Period $T = 12\ \text{hours} = 12 \times 3600\ \text{seconds}$.
• Frequency:

$f = \frac{1}{12 \times 3600}Hz$

4. Option Matching

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. Amplitude: $3 \ \text{m}$, Frequency: $\frac{3600}{12}Hz$ — Unit Conversion Error
  Inverting the frequency formula incorrectly.
• C. Amplitude: $6 \ \text{m}$, Frequency: $\frac{1}{24 \times 3600}Hz$ — Unit Conversion Error
  Confusing amplitude with the total peak-to-peak depth variation (6 m).
• D. Amplitude: $6 \ \text{m}$, Frequency: $\frac{3600}{24}Hz$ — Unit Conversion Error
  Using the wrong amplitude and inverting the frequency formula.
• E. Amplitude: $8 \ \text{m}$, Frequency: $\frac{1}{12 \times 3600}Hz$ — Unit Conversion Error
  Using the maximum depth instead of the amplitude.
• F. Amplitude: $8 \ \text{m}$, Frequency: $\frac{3600}{12}Hz$ — Unit Conversion Error
  Using the maximum depth and inverting the frequency.
• G. Amplitude: $16 \ \text{m}$, Frequency: $\frac{1}{24 \times 3600}Hz$ — Unit Conversion Error
  Using the peak depth and assuming a 24 hour period.
• H. Amplitude: $16 \ \text{m}$, Frequency: $\frac{3600}{24}Hz$ — Unit Conversion Error
  Using the peak depth and inverting a 24 hour period.

Common Mistake (⚠️):
Confusing amplitude with peak-to-peak depth variation ($6 \ \text{m}$).

Takeaway (📌):
Amplitude is half the total range from minimum to maximum.

Correct Answer: C. Atomic number 82, Mass number 207

Fastest Approach (🚀):
Mass number: $219 - 3(4) = 207$.
Atomic number: $86 - 3(2) + 2(1) = 82$.

Matches Option C.

Step-by-Step Breakdown:

1. Nuclear Decay Transformation Rules

A. Alpha Decay ($\alpha = {}^4_2 \text{He}$)

$\Delta A = -4, \quad \Delta Z = -2$

B. Beta-Minus Decay ($\beta^- = {}^{\ 0}_{-1}\text{e}$)

$\Delta A = 0, \quad \Delta Z = +1$

2. Initial Isotope Coordinates

• Starting isotope: Radon-219 ($_{86}^{219}\text{Rn}$).
• Initial mass number $A_i = 219$.
• Initial atomic number $Z_i = 86$.

3. Decay Chain Arithmetic

• Decay process: $3\alpha + 2\beta^-$.
• Mass number calculation:
• Atomic number calculation:

$A_f = 219 - 3(4) - 2(0) = 219 - 12 = 207$
$Z_f = 86 - 3(2) + 2(1) = 86 - 6 + 2 = 82$

4. Option Matching

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. Atomic number 80, Mass number 207 — Sign Error
  Forgetting that beta-minus emission increases the atomic number, rather than decreasing it.
• B. Atomic number 80, Mass number 211 — Incomplete Calculation
  Miscalculating the mass number change from the alpha particles.
• D. Atomic number 82, Mass number 215 — Unit Conversion Error
  Calculating the mass change for only a single alpha particle decay.
• E. Atomic number 85, Mass number 211 — Sign Error
  Making a sign error when adjusting the atomic number.
• F. Atomic number 85, Mass number 219 — Conceptual Misunderstanding
  Assuming no mass change occurred during the decays.
• G. Atomic number 86, Mass number 215 — Sign Error
  Making a sign error when adjusting the atomic number.
• H. Atomic number 86, Mass number 219 — Conceptual Misunderstanding
  Forgetting to change the atomic and mass numbers altogether.

Common Mistake (⚠️):
Forgetting that beta-minus emission increases the atomic number by $1$, leading to an incorrect atomic number of $80$ (Option A).

Takeaway (📌):
Keep track of charge and mass conservation. Alpha: $\Delta A = -4, \Delta Z = -2$. Beta-minus: $\Delta A = 0, \Delta Z = +1$.

Correct Answer: B. Bulb X: Brighter, Bulb Y: Dimmer

Fastest Approach (🚀):
Initial state (P open, Q closed): $R_{\text{total}} = 2.5R \implies I_y= 0.4 \ \text{V/R}, I_x= 0.2 \ \text{V/R}$.
Final state (P closed, Q open): Bottom branch open-circuited $\implies R_{\text{total}} = 3R \implies I_y= I_x= 0.33 \ \text{V/R}$.
Bulb X: $0.2 \to 0.33 \implies$ Brighter.
Bulb Y: $0.4 \to 0.33 \implies$ Dimmer.

Matches Option B.

Step-by-Step Breakdown:

1. DC Circuit Laws & Filament Bulb Brightness

A. Bulb Brightness Law

Dissipated power $P = I^2 R$. Brightness depends directly on current magnitude $I$.

B. Series & Parallel Equivalent Resistance

$R_{\text{series}} = R_1 + R_2, \quad R_{\text{parallel}} = \frac{R_1 R_2}{R_1 + R_2}$

2. Initial Circuit Analysis (P Open, Q Closed)

• Parallel sub-network 1 (Bulb X || bottom-left): $R_{p1} = 0.5R$.
• Sub-network 2 (bottom-middle): $R_{p2} = R$.
• Series combination of the two sub-networks: $R_p = 0.5R + R = 1.5R$.
• Total circuit resistance: $R_{\text{initial}} = 1.5R + R = 2.5R$.
• Total current (through Bulb Y): $I_{Y, \text{initial}} = \frac{V}{2.5R} = 0.4 \frac{V}{R}$.
• Voltage drop across $R_{p1} = 0.4 \frac{V}{R} \times 0.5R = 0.2 \ \text{V}$.
• Current through Bulb X: $I_{X, \text{initial}} = \frac{0.2 \ \text{V}}{R} = 0.2 \frac{V}{R}$.

3. Final Circuit Analysis (P Closed, Q Open)

• Q open $\implies$ bottom wire open-circuited (disabling both bottom bulbs).
• P closed $\implies$ top row (Bulb X + top-middle bulb) forms series branch of $2R$.
• Total circuit resistance: $R_{\text{final}} = 2R + R = 3R$.
• Total current (through BOTH X and Y): $I_{X, \text{final}} = I_{Y, \text{final}} = \frac{V}{3R} \approx 0.333 \frac{V}{R}$.

4. Brightness Comparison & Option Matching

• Bulb X: $0.20 \frac{V}{R} \to 0.333 \frac{V}{R} \implies$ Brighter.
• Bulb Y: $0.40 \frac{V}{R} \to 0.333 \frac{V}{R} \implies$ Dimmer.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. Bulb X: Brighter, Bulb Y: Brighter — Conceptual Misunderstanding
  Incorrectly determining the final current distribution.
• C. Bulb X: Dimmer, Bulb Y: Brighter — Unit Conversion Error
  Miscalculating the equivalent resistance and assuming Bulb X gets less current.
• D. Bulb X: Dimmer, Bulb Y: Dimmer — Conceptual Misunderstanding
  Incorrectly assuming both bulbs receive less current due to the open switch.
• E. Bulb X: Unchanged, Bulb Y: Brighter — Conceptual Misunderstanding
  Assuming the series Bulb X doesn't change because the battery voltage is constant.
• F. Bulb X: Brighter, Bulb Y: Unchanged — Conceptual Misunderstanding
  Assuming Bulb Y's brightness doesn't change.

Common Mistake (⚠️):
Assuming the total current is constant, or failing to realize that opening Switch Q open-circuits the entire bottom branch.

Takeaway (📌):
Calculate equivalent resistances for both states to determine current changes through specific components.

Correct Answer: G. Power: $250 \ \text{W}$, Speed: $10 \ \text{m/s}$

Fastest Approach (🚀):
Speed: $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 5} = 10 \ \text{m/s}$.
Power: $P = \frac{1}{2} \left(\frac{dm}{dt}\right) v^2 = 0.5 \times 5 \times 100 = 250 \ \text{W}$.

Matches Option G.

Step-by-Step Breakdown:

1. Conservation of Energy & Fluid Kinetic Power Principles

A. Torricelli Energy Conservation

$\frac{1}{2} m v^2 = m g h \implies v = \sqrt{2gh}$

B. Mass Flow Kinetic Power Formula

$P = \frac{d E_k}{dt} = \frac{1}{2} \left(\frac{dm}{dt}\right) v^2$

2. Leaving Speed Calculation

• Maximum height $h = 5 \ \text{m}, g = 10 \text{N/kg}$.
• Exit speed: $v = \sqrt{2 \times 10 \times 5} = 10 \ \text{m/s}$.

3. Pump Power Calculation

• Mass flow rate $\dot{m} = 5\ \text{kg/s}$.
• Pump power output:

$P = \frac{1}{2} \times (5\ \text{kg/s}) \times (10 \ \text{m/s})^2 = 2.5 \times 100 = 250 \ \text{W}$

4. Option Matching

Matches Option G.

Why the Other Options Are Wrong (❌):

• A. Power: $5 \ \text{W}$, Speed: $5 \ \text{m/s}$ — Misapplied Formula
  Miscalculating both the exit speed and power.
• B. Power: $5 \ \text{W}$, Speed: $50 \ \text{m/s}$ — Incomplete Calculation
  Making an arithmetic error in the speed calculation.
• C. Power: $25 \ \text{W}$, Speed: $10 \ \text{m/s}$ — Misapplied Formula
  Calculating the correct speed but making a factor of 10 error on power.
• D. Power: $25 \ \text{W}$, Speed: $100 \ \text{m/s}$ — Unit Conversion Error
  Using the wrong power formula.
• E. Power: $50 \ \text{W}$, Speed: $5 \ \text{m/s}$ — Unit Conversion Error
  Using the wrong power formula.
• F. Power: $50 \ \text{W}$, Speed: $50 \ \text{m/s}$ — Unit Conversion Error
  Using the wrong power formula.
• H. Power: $250 \ \text{W}$, Speed: $100 \ \text{m/s}$ — Conceptual Misunderstanding
  Making a factor of 10 error on the speed calculation.

Common Mistake (⚠️):
Calculating power as $mgh = 5 \times 10 \times 5 = 250$ W but getting the speed formula wrong, or using $P = mgh$ directly without confirming the kinetic energy transfer rate.

Takeaway (📌):
Use conservation of energy to link velocity and height ($v = \sqrt{2gh}$), and power to the rate of kinetic energy delivery ($P = \frac{1}{2} \dot{m} v^2$).

Correct Answer: D. 45 kJ

Fastest Approach (🚀):
Parallel gravity force: $mg \sin(\theta) = 800 \times 10 \times \frac{1}{20} = 400 \ \text{N}$.
Frictional force: $500 \ \text{N}$.
Total engine force: $400 + 500 = 900 \ \text{N}$.
Work done: $W = F \times d = 900 \times 50 = 45,000 \text{J} = 45 \text{kJ}$.

Matches Option D.

Step-by-Step Breakdown:

1. Incline Geometry & Newton's First Law Principles

A. Incline Ratio Rule

Slope ratio $1\ \text{in }20 \implies \sin\theta = \frac{1}{20}$.

B. Gravitational Slope Force

$F_{g\parallel} = mg \sin\theta$

C. Constant Speed Equilibrium

$\sum F = 0 \implies F_{\text{engine}} = F_{g\parallel} + f_{\text{friction}}$

2. Opposing Force Summation

• Mass $m = 800 \ \text{kg}, g = 10 \text{N/kg}, \sin\theta = \frac{1}{20}$.
• Parallel gravity force: $F_{g\parallel} = 800 \times 10 \times \frac{1}{20} = 400 \ \text{N}$.
• Friction: $f = 500 \ \text{N}$.
• Engine force required: $F_{\text{engine}} = 400 + 500 = 900 \ \text{N}$.

3. Work Calculation

Distance $d = 50 \ \text{m}$.
$W = F_{\text{engine}} \times d = 900 \times 50 = 45,000\ \text{J} = 45 \text{kJ}$

4. Option Matching

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. 20 kJ — Unit Conversion Error
  Forgetting to include the frictional force opposing motion.
• B. 25 kJ — Unit Conversion Error
  Forgetting to include the gravitational force opposing motion.
• C. 27 kJ — Incomplete Calculation
  Making an arithmetic error when summing the forces.
• E. 65 kJ — Conceptual Misunderstanding
  Miscalculating the gravitational component.
• F. 160 kJ — Unit Conversion Error
  Using cosine instead of sine for the slope.

Common Mistake (⚠️):
Forgetting the frictional force ($500$ N), which would give $W = 400 \times 50 = 20$ kJ (Option A).

Takeaway (📌):
Ensure all opposing forces (both gravity component and friction) are summed when calculating the required engine force for constant-speed motion.

Correct Answer: B. $3.75 \times 10^3 \ \text{N}$

Fastest Approach (🚀):
Initial kinetic energy: $E_k = 0.5 \times 0.05 \times 300^2 = 2250 \ \text{J}$.
Stopping force: $F = \frac{E_k}{s} = \frac{2250}{0.6} = 3750 \text{N} = 3.75 \times 10^3 \ \text{N}$.

Matches Option B.

Step-by-Step Breakdown:

1. Work-Energy Theorem & SI Unit Laws

A. Kinetic Energy Formula

$K = \frac{1}{2} m u^2$

B. Work-Energy Theorem

$W_{\text{net}} = \Delta K \implies F \cdot s = \frac{1}{2} m u^2$

2. Initial Kinetic Energy Calculation

• $m = 50\ \text{g} = 0.05 \ \text{kg}, u = 300 \ \text{m/s}$.
• $K = \frac{1}{2} (0.05) (300)^2 = 0.025 \times 90,000 = 2250 \ \text{J}$.

3. Average Stopping Force Calculation

• Penetration distance $s = 60 \ \text{cm} = 0.60 \ \text{m}$.
• $F \times 0.60 = 2250 \implies F = \frac{2250}{0.60} = 3750 \text{N} = 3.75 \times 10^3 \ \text{N}$.

4. Option Matching

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $37.5 \ \text{N}$ — Conceptual Misunderstanding
  Failing to convert grams to kilograms, resulting in an error of 1000.
• C. $3.75 \times 10^4 \ \text{N}$ — Conceptual Misunderstanding
  Making an order of magnitude error with the decimal point.
• D. $3.75 \times 10^6 \ \text{N}$ — Conceptual Misunderstanding
  Failing to convert the stopping distance from cm to m.
• E. $3.75 \times 10^5 \ \text{N}$ — Incomplete Calculation
  Squaring the mass or making another arithmetic error.

Common Mistake (⚠️):
Failing to convert units (like using mass in grams or distance in centimetres), which leads to an error of several orders of magnitude.

Takeaway (📌):
Always perform calculations in SI base units (kg, m, s) to ensure scientific notation prefixes are correct.

Correct Answer: C. $\frac{5}{3}V$

Fastest Approach (🚀):
$V_{\text{battery}} = V \times \frac{3R + 2R}{3R} = \frac{5}{3}V$.

Matches Option C.

Step-by-Step Breakdown:

1. Potential Divider Theorem & Series Resistors

A. Series Current Law

Same current $I$ flows through all components in series.

B. Potential Divider Relation

$\frac{V_1}{V_{\text{total}}} = \frac{R_1}{R_{\text{total}}}$

2. Resistance & Current Formulation

• Measured resistor $R_1 = 3R$, second resistor $R_2 = 2R$.
• Total series resistance: $R_{\text{total}} = 3R + 2R = 5R$.
• Circuit current: $I = \frac{V}{3R}$.

3. Total Battery Voltage Calculation

$V_{\text{battery}} = I \cdot R_{\text{total}} = \left(\frac{V}{3R}\right) \cdot (5R) = \frac{5}{3} V$

4. Option Matching

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $\frac{8}{3}V$ — Misapplied Formula
  Setting up the ratio incorrectly.
• B. $5 \ \text{V}$ — Incomplete Calculation
  Dropping the fractional part.
• D. $\frac{3}{5}V$ — Unit Conversion Error
  Setting up the potential divider ratio to find the voltage across the 3R resistor, rather than solving for the battery voltage.
• E. $2 \ \text{V}$ — Incomplete Calculation
  An arithmetic error in the fraction.

Common Mistake (⚠️):
Calculating $\frac{3}{5}V$ (Option D) instead of solving for battery voltage.

Takeaway (📌):
In series circuits, battery voltage is $V_{\text{total}} = V_{\text{measured}} \times \frac{R_{\text{total}}}{R_{\text{measured}}}$.

Correct Answer: D. I and II

Fastest Approach (🚀):
Speed depends on medium $\implies$ equal speeds (III false).
$\lambda_A = \frac{1}{2} \lambda_B \implies f_A = 2 f_B \implies$ higher pitch (II true).
Intensity $I \propto A^2 f^2$. Same amplitude, but $f_A = 2f_B \implies I_A = A^2(2f_B)^2 = 4 A^2 f_B^2 = 4 I_B$, so Wave A's intensity IS greater (I true).

Matches Option D (I and II).

Step-by-Step Breakdown:

1. Wave Propagation Physics Laws

A. Medium Invariance of Speed

Propagation speed $v$ is determined strictly by physical medium properties. $v_A = v_B$ in the same uniform medium.

B. Wave Relation & Pitch

$v = f \lambda \implies f = \frac{v}{\lambda}$
Human pitch perception depends directly on frequency $f$.

C. Intensity Law

For a mechanical/sound wave, intensity is $I = \frac{1}{2}\rho v \omega^2 A^2 \propto A^2 f^2$ (not amplitude alone). Intensity depends on the SQUARE of amplitude AND the SQUARE of frequency.

2. Analysis of Statements

• Statement III (Speed): Same uniform medium $\implies v_A = v_B \implies$ INCORRECT.
• Statement II (Pitch): $\lambda_A = \frac{1}{2} \lambda_B$ and $v_A = v_B \implies f_A = \frac{v_A}{\lambda_A} = \frac{v_B}{\lambda_B/2} = 2 f_B \implies$ higher pitch $\implies$ CORRECT.
• Statement I (Intensity): $I \propto A^2 f^2$. Amplitudes are equal, but $f_A = 2f_B$, so $\frac{I_A}{I_B} = \left(\frac{A_A}{A_B}\right)^2\left(\frac{f_A}{f_B}\right)^2 = 1^2 \times 2^2 = 4$. Wave A's intensity is 4x that of Wave B $\implies$ CORRECT.

3. Option Matching

Statements I and II are correct; III is false.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. Only I — Incomplete Selection
  Statement III is also needed: Statement II (pitch) is also correct, since Wave A's frequency is double Wave B's, so 'Only I' misses that.
• B. Only II — Conceptual Misunderstanding
  Assumes intensity depends only on amplitude, ignoring that intensity also scales with frequency squared ($I \propto A^2 f^2$). Since Wave A has double the frequency, its intensity is actually 4x greater, so Statement I is also true.
• C. Only III — Conceptual Misunderstanding
  The speeds are exactly equal because the medium is identical; Statement III is false, and this option also misses the two true statements (I and II).
• E. II and III — Conceptual Misunderstanding
  Statement III is false because wave speed is invariant in a uniform medium; this option incorrectly includes III and omits I.

Common Mistake (⚠️):
Believing intensity depends on amplitude ALONE (a common oversimplification). The correct relationship, $I \propto A^2 f^2$, means two waves of equal amplitude but different frequency do NOT have equal intensity. A student who stops at 'equal amplitude therefore equal intensity' will incorrectly mark Statement I as false and select Option B instead of D.

Takeaway (📌):
Wave speed depends on medium. Pitch depends on frequency. Intensity depends on BOTH amplitude and frequency ($I \propto A^2 f^2$), not amplitude alone.

Correct Answer: C. 3Q/2

Fastest Approach (🚀):
For object A: $Q = 2 \ \text{m} \cdot c \cdot T = 2mcT$.
For object B: $Q_B = m \cdot 3c \cdot T = 3mcT$.
Ratio: $\frac{Q_B}{Q} = \frac{3mcT}{2mcT} = 1.5 \implies Q_B = \frac{3}{2}Q$.

Matches Option C.

Step-by-Step Breakdown:

1. Specific Heat Capacity Equation

The fundamental equation linking heat energy, mass, specific heat capacity, and temperature change is:
$Q = m \cdot c \cdot \Delta T$
Where:

• $Q$ is the heat energy transferred
• $m$ is the mass of the object
• $c$ is the specific heat capacity (energy required to raise 1 kg by 1 Kelvin)
• $\Delta T$ is the change in temperature

2. analysing Object A

• The mass is $m_A = 2m$.
• The specific heat capacity is $c_A = c$.
• The temperature changes from $T$ to $2T$, so $\Delta T = 2T - T = T$.

The heat received by object A is defined as $Q$.
Using our formula:
$Q_A = (2m) \cdot (c) \cdot (T) = 2mcT$
So we know that $Q = 2mcT$. We can rearrange this to find a useful substitution: $mcT = \frac{Q}{2}$.

3. analysing Object B

• The mass is $m_B = m$.
• The specific heat capacity is $c_B = 3c$.
• The temperature change is identical, $\Delta T = T$.

The heat received by object B is $Q_B$.
$Q_B = (m) \cdot (3c) \cdot (T) = 3mcT$

4. Final Ratio & Substitution

We need to express $Q_B$ in terms of the original $Q$.
Substitute our rearranged term $mcT = \frac{Q}{2}$ directly into the equation for $Q_B$:
$Q_B = 3(mcT) = 3 \left(\frac{Q}{2}\right) = \frac{3}{2} Q$

5. Option Matching

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. Q — Incomplete Calculation
  Forgetting to account for the differences in mass and specific heat between the two objects.
• B. 2Q/3 — Unit Conversion Error
  Inverting the final mathematical ratio.
• D. 2Q — Incomplete Calculation
  Making an arithmetic error during the substitution step.
• E. 3Q — Conceptual Misunderstanding
  Only comparing the specific heat capacities (3:1) and entirely ignoring the mass difference (1:2).

Common Mistake (⚠️):
Dividing specific heat values incorrectly, or only looking at the specific heat capacity ratio (3:1) and entirely forgetting to account for the difference in mass (1:2).

Takeaway (📌):
When comparing two thermal systems, calculate individual products of mass, specific heat, and temperature change for both, then find the relative ratio of their heat energy.

Correct Answer: D. II, III, IV

Fastest Approach (🚀):
After $\alpha$ emission ($Z-2, A-4$), atomic number becomes $90$. After $\beta^-$ emission ($Z+1$), atomic number becomes $91$. Statement I is false.
Gamma rays are EM waves (II true). Decays release energy (III true). Gamma rays have greatest range in air (IV true).
Statements II, III, IV are correct.

Matches Option D.

Step-by-Step Breakdown:

1. Nuclear Transformation & Decay Particle Characteristics

• Alpha Emission: Helium nucleus ($^4_2 \text{He}$), $\Delta Z = -2, \Delta A = -4$.
• Beta-Minus Emission: Fast electron ($^0_{-1}\text{e}$), $\Delta Z = +1, \Delta A = 0$.
• Gamma Emission: High-frequency photon (EM wave), no mass or charge, high penetration/range.

2. Statement-by-Statement Verification

• Statement I (Turns into X Nucleus): Starting $Z, A$. After $\alpha$, $Z-2$. After $\beta$, $Z-2+1 = Z-1 ≠ Z \implies$ FALSE.
• Statement II (Electromagnetic Waves Emitted): Gamma rays are EM waves $\implies$ TRUE.
• Statement III (Nuclear Energy Decreases): Spontaneous decay releases energy to lower, more stable nuclear state $\implies$ TRUE.
• Statement IV (Gamma Greatest Range in Air): Neutral gamma photons interact less frequently with air molecules than charged $\alpha$ or $\beta$ particles $\implies$ greatest range $\implies$ TRUE.

3. Option Matching

Statements II, III, IV are correct.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. III and IV — Unit Conversion Error
  Missing Statement II, which is also a scientifically correct statement.
• B. I and III — Conceptual Misunderstanding
  Statement I is false because beta decay changes the atomic number back to Z-1, not the original Z.
• C. I, II and III — Conceptual Misunderstanding
  Statement I is false for the reason above.
• E. I, II, III, and IV — Conceptual Misunderstanding
  Statement I is false for the reason above.

Common Mistake (⚠️):
Failing to apply the beta decay step to the atomic number, thereby assuming Statement I is correct.

Takeaway (📌):
analyse radioactive sequences step-by-step. Beta decay increases atomic number, which changes the chemical element.

Correct Answer: A. $-1000 \ \text{m/s}^2$

Fastest Approach (🚀):
$u = 108 \ \text{km/h} = 30 \ \text{m/s}, s = 0.45 \ \text{m}$.
$a = -\frac{u^2}{2 \ \text{s}} = -\frac{900}{0.9} = -1000 \text{m s}^{-2}$.

Matches Option A.

Step-by-Step Breakdown:

1. Standardizing Units (Crucial First Step)

Before applying any kinematic equations, all values must be in standard SI units (metres and seconds). The initial speed $u$ is given in $\text{km/h}$.
To convert from $\text{km/h}$ to $\text{m/s}$, we divide by 3.6 (since there are 1000 metres in a kilometre and 3600 seconds in an hour, $1000/3600 = 1/3.6$).
$u = 108 \ \text{km/h} = \frac{108}{3.6} = 30 \ \text{m/s}$

2. Selecting the Correct SUVAT Equation

We need to find the constant deceleration ($a$). We know:

• Initial velocity $u = 30 \ \text{m/s}$
• Final velocity $v = 0 \ \text{m/s}$ (since the car comes to a stop)
• Displacement $s = 0.45 \ \text{m}$ (the compression distance of the car parts)
• Time $t$ is unknown and unneeded.

Because we have $u, v, s$ and want $a$, we use the time-independent kinematic equation:
$v^2 = u^2 + 2 a s$

3. Deceleration Calculation

Substitute the known values into the equation:
$0^2 = (30)^2 + 2 \cdot a \cdot (0.45)$
$0 = 900 + 0.9a$
Rearrange to solve for $a$:
$0.9a = -900$
$a = -\frac{900}{0.9}$
$a = -1000 \text{m s}^{-2}$
The negative sign perfectly indicates deceleration, exactly as expected for a collision.

4. Option Matching

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $-100 \ \text{m/s}^2$ — Unit Conversion Error
  Using an incorrect speed conversion factor or making a decimal arithmetic error.
• C. $-10 \ \text{m/s}^2$ — Unit Conversion Error
  Making an order of magnitude error during the final division.
• D. $-900 \ \text{m/s}^2$ — Misapplied Formula
  Forgetting the factor of 2 in the kinematic equation (v^2 = u^2 + as).
• E. $-90 \ \text{m/s}^2$ — Conceptual Misunderstanding
  Forgetting the factor of 2 and making a decimal error.

Common Mistake (⚠️):
Plugging 108 directly into the SUVAT equation without converting it from km/h to m/s, which results in a wildly incorrect deceleration of $-12960 \text{m/s}^2$.

Takeaway (📌):
Always verify that your speed units are in standard SI metres per second (m/s) before applying equations of motion. If they are in km/h, divide by 3.6.