ESAT Paper 2 Physics Worked Solutions

Correct Answer: A. $1.25 \ \text{kg}$

Fastest Approach (🚀):
Use $E_{k} = \frac{p^2}{2m} \implies m = \frac{p^2}{2 E_{k}} = \frac{50^2}{2000} = 1.25 \ \text{kg}.$

Matches Option A.

Step-by-Step Breakdown:

1. Combining Momentum and Kinetic Energy Formulas

Recall the standard definitions of linear momentum ($p$) and kinetic energy ($E_k$):

• $p = m v \implies v = \frac{p}{m}$
• $E_k = \frac{1}{2} m v^2$

Substitute $v = \frac{p}{m}$ into the kinetic energy equation:
$E_k = \frac{1}{2} m \left(\frac{p}{m}\right)^2 = \frac{1}{2} m \frac{p^2}{m^2} = \frac{p^2}{2m}$
This yields the fundamental relation between kinetic energy, momentum, and mass:
$E_k = \frac{p^2}{2m}$
Rearranging to solve directly for mass ($m$):
$m = \frac{p^2}{2 E_k}$

2. Parameter Conversion & Calculation

• Momentum $p = 50 \ \text{N s}$
• Kinetic Energy $E_k = 1 \ \text{kJ} = 1000 \ \text{J}$

Substitute the known values into our mass equation:
$m = \frac{50^2}{2 \times 1000} = \frac{2500}{2000}$
Simplify the fraction by dividing top and bottom by 500:
$m = \frac{5}{4} = 1.25 \ \text{kg}$

3. Option Matching

Matches Option A.

Common Mistake (⚠️):
First solving for velocity $v$ and then substituting back into $p = mv$, which requires two full algebraic steps instead of one direct substitution.

Takeaway (📌):
Always memorize the relationship $E_k = \frac{p^2}{2m}$—it is one of the most useful time-saving shortcuts in Physics admissions tests!

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

Fastest Approach (🚀):
Equivalent resistance of two identical $1 \ \text{k}\Omega$ resistors in parallel is $500\,\Omega.$
Total current $I = \frac{5 \ \text{V}}{500\,\Omega} = 0.01 \ \text{A} = 10 \ \text{mA}.$

Matches Option D.

Step-by-Step Breakdown:

1. Circuit Analysis & Ammeter Placement

The circuit diagram consists of a $5 \ \text{V}$ ideal DC voltage source connected to two identical resistors in parallel, each with resistance $R = 1 \ \text{k}\Omega = 1000 \ \Omega$.
Notice that the ammeter is connected directly in series with the main power rail before the current splits into the two parallel branches.
Therefore, the ammeter measures the total circuit current ($I_{\text{total}}$) leaving the battery.

2. Calculating Equivalent Parallel Resistance

For two resistors in parallel, the total equivalent resistance $R_{\text{eq}}$ is given by:
$\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2}$
Since the two parallel resistors are identical ($R_1 = R_2 = 1000 \ \Omega$), the equivalent resistance is simply half the value of a single resistor:
$R_{\text{eq}} = \frac{R}{2} = \frac{1000 \ \Omega}{2} = 500 \ \Omega$

3. Applying Ohm's Law

Using Ohm's Law ($I = \frac{V}{R}$) for the entire circuit:
$I_{\text{total}} = \frac{V_{\text{battery}}}{R_{\text{eq}}} = \frac{5 \ \text{V}}{500 \ \Omega}$
$I_{\text{total}} = \frac{1}{100} \ \text{A} = 0.01 \ \text{A}$

4. Unit Conversion to Milliamperes

Convert amperes ($\text{A}$) to milliamperes ($\text{mA}$) by multiplying by 1000:
$I_{\text{total}} = 0.01 \times 1000 = 10 \ \text{mA}$

5. Option Matching

Matches Option D.

Common Mistake (⚠️):
Forgetting to convert $1 \ \text{k}\Omega$ to $1000 \ \Omega$ before applying Ohm's Law, or miscalculating the branch current ($5 \ \text{mA}$) as the main line current.

Takeaway (📌):
When two identical resistors of value $R$ are connected in parallel, their total equivalent resistance is instantly $R/2$.

Correct Answer: C. $0.4 \ \text{ms}$

Fastest Approach (🚀):
Double depth for round trip: $s = 2 \times 0.08 = 0.16 \ \text{m}.$
Time $t = \frac{0.16}{400} = 0.0004 \ \text{s} = 0.4 \ \text{ms}.$

Matches Option C.

Step-by-Step Breakdown:

1. Identifying Distractor Data & Echo Kinematics

• Distractor Information: The frequency of the ultrasound wave ($f = 60 \ \text{kHz}$) is not needed to calculate travel time. The speed of sound in tissue ($v = 400 \ \text{m/s}$) is independent of frequency in a non-dispersive medium.
• Echo Principle: An ultrasound pulse travels from the transmitter on the skin surface down to the foetus at depth $d$, reflects, and travels back up to the receiver on the skin surface.

Therefore, the total distance traveled by the pulse is double the depth:
$s = 2d = 2 \times (8 \ \text{cm}) = 16 \ \text{cm}$

2. SI Unit Conversions

Convert distance $s$ from centimetres to metres:
$s = 16 \ \text{cm} = 0.16 \ \text{m}$
Speed of sound $v = 400 \ \text{m/s}$.

3. Time Calculation

Using the uniform velocity formula $s = v \cdot t \implies t = \frac{s}{v}$:
$t = \frac{0.16 \ \text{m}}{400 \ \text{m/s}}$
Simplify the division:
$t = \frac{16 \times 10^{-2}}{4 \times 10^2} = \left(\frac{16}{4}\right) \times 10^{-2 - 2} = 4 \times 10^{-4} \ \text{seconds}$

4. Unit Conversion to Milliseconds

Convert seconds to milliseconds ($1 \ \text{ms} = 10^{-3} \ \text{s}$):
$t = 4 \times 10^{-4} \ \text{s} = 0.4 \times 10^{-3} \ \text{s} = 0.4 \ \text{ms}$

5. Option Matching

Matches Option C.

Common Mistake (⚠️):
Forgetting that ultrasound detection uses echoes, and calculating the time for a one-way trip ($0.08 / 400 = 0.2 \ \text{ms}$), which leads directly to trap Option B.

Takeaway (📌):
Whenever a physics problem involves reflection, radar, sonar, or ultrasound echo-ranging, ALWAYS double the target distance before calculating time!

Correct Answer: B. $17.5 \ \text{m}$

Fastest Approach (🚀):
Triangle 1 ($0 \to 2 \ \text{s}$): $\frac{1}{2} \times 2 \times 5 = 5 \ \text{m}.$
Triangle 2 ($2 \to 7 \ \text{s}$): $\frac{1}{2} \times 5 \times 5 = 12.5 \ \text{m}.$
Total $= 5 + 12.5 = 17.5 \ \text{m}.$

Matches Calculated value $17.5 \ \text{m}.$

Matches Option B.

Step-by-Step Breakdown:

1. Velocity-Time Graph Calculus Principles

On a velocity-time ($v$-$t$) graph:

• The gradient represents acceleration ($a = \frac{\text{d}v}{\text{d}t}$).
• The area under the curve represents total distance / displacement ($s = \int v \, \text{d}t$).

2. Geometric Decomposition of the Motion

The motion consists of two distinct triangular phases:

• Phase 1 (Acceleration, $t = 0$ to $t = 2 \ \text{s}$):

A right-angled triangle with base $b_1 = 2 - 0 = 2 \ \text{s}$ and peak height $h_1 = 5 \ \text{m/s}$.
$\text{Area}_1 = \frac{1}{2} \times b_1 \times h_1 = \frac{1}{2} \times 2 \times 5 = 5 \ \text{m}$

• Phase 2 (Deceleration, $t = 2$ to $t = 7 \ \text{s}$):

A right-angled triangle with base $b_2 = 7 - 2 = 5 \ \text{s}$ and peak height $h_2 = 5 \ \text{m/s}$.
$\text{Area}_2 = \frac{1}{2} \times b_2 \times h_2 = \frac{1}{2} \times 5 \times 5 = 12.5 \ \text{m}$

3. Summing Total Distance

Sum the areas of both geometric regions:
$\text{Total Distance} = \text{Area}_1 + \text{Area}_2 = 5 \ \text{m} + 12.5 \ \text{m} = 17.5 \ \text{m}$

4. Option Matching

Calculated exact geometric distance is $17.5 \ \text{m}$.

Matches Option B.

Common Mistake (⚠️):
Misreading the base of the second triangle as 7 instead of $(7 - 2) = 5$, or forgetting the factor of $1/2$ for triangle areas.

Takeaway (📌):
Always compute the base width of a section as $\Delta t = t_{\text{final}} - t_{\text{initial}}$ rather than using the absolute clock time reading.

Correct Answer: A. Alpha emission

Fastest Approach (🚀):
Mass number drops by 4 ($210 \to 206$).
Atomic number drops by 2 ($86 \to 84$).
This is an alpha particle ($ {}_2^4 \text{He}$).

Matches Option A.

Step-by-Step Breakdown:

1. Nuclear Conservation Principles

In all nuclear reactions and radioactive decay processes, two fundamental quantities are strictly conserved across both sides of the nuclear equation:

1. Total Mass Number ($A$): Sum of protons and neutrons (top index).
2. Total Atomic Number ($Z$): Total electric charge / proton count (bottom index).

2. Balancing the Nuclear Equation

The given decay reaction is:
$ {}_{86}^{210}\text{Rn} \rightarrow {}_{84}^{206}\text{Po} + {}_{Z}^A X$

• Conservation of Mass Number ($A$):
• Conservation of Atomic Number ($Z$):

$210 = 206 + A \implies A = 210 - 206 = 4$
$86 = 84 + Z \implies Z = 86 - 84 = 2$

3. Emitted Particle Identification

The emitted particle has mass number $A = 4$ and atomic number $Z = 2$.
This particle consists of 2 protons and 2 neutrons, which is a Helium-4 nucleus ($ {}_2^4 \text{He}$ or $\alpha$).
Therefore, the decay process is Alpha emission.

4. Option Matching

Matches Option A.

Common Mistake (⚠️):
Confusing alpha and beta decay. Beta-minus decay leaves the mass number $A$ completely unchanged while increasing atomic number $Z$ by 1.

Takeaway (📌):
Memorize the signature decay patterns: Alpha decay drops mass number by 4 and atomic number by 2; Beta decay leaves mass number unchanged.

Correct Answer: A. $T=\frac{P}{Lf}$

Fastest Approach (🚀):
Units of Force $T = kg\cdot m\cdot s^{-2}.$
Evaluate $\frac{P}{Lf} = \frac{kg\cdot m^2\cdot s^{-3}}{m\cdot s^{-1}} = kg\cdot m\cdot s^{-2}.$
Matches force units.

Matches Option A.

Step-by-Step Breakdown:

1. Fundamentals of Dimensional Analysis

Every physically valid equation must be dimensionally homogeneous, meaning the fundamental base units (mass $\text{kg}$, length $\text{m}$, time $\text{s}$) on the left-hand side must equal the base units on the right-hand side.

Let's determine the base SI units for each variable in the problem:

• Tension Force ($T$): $[T] = \text{Force} = \text{mass} \times \text{acceleration} = \text{kg} \cdot \text{m} \cdot \text{s}^{-2}$
• Power ($P$): $[P] = \frac{\text{Energy}}{\text{time}} = \frac{\text{Force} \times \text{distance}}{\text{time}} = \frac{(\text{kg} \cdot \text{m} \cdot \text{s}^{-2}) \cdot \text{m}}{\text{s}} = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-3}$
• Length ($L$): $[L] = \text{m}$
• Frequency ($f$): $[f] = \frac{1}{\text{time}} = \text{s}^{-1}$

2. Testing Candidate Expressions for Force Dimensions

We need to find an expression among the options whose base units evaluate to $\text{kg} \cdot \text{m} \cdot \text{s}^{-2}$.

Let's evaluate the expression $\frac{P}{L f}$:
$\left[ \frac{P}{L f} \right] = \frac{[P]}{[L] [f]} = \frac{\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-3}}{\text{m} \cdot \text{s}^{-1}}$
Simplify the metres ($\text{m}$) term:
$\frac{\text{m}^2}{\text{m}} = \text{m}^1 = \text{m}$
Simplify the seconds ($\text{s}$) term:
$\frac{\text{s}^{-3}}{\text{s}^{-1}} = \text{s}^{-3 - (-1)} = \text{s}^{-2}$
Combining these gives:
$\left[ \frac{P}{L f} \right] = \text{kg} \cdot \text{m} \cdot \text{s}^{-2}$
This matches the base SI units of Force ($T$) perfectly.

3. Option Matching

Matches Option A.

Common Mistake (⚠️):
Trying to derive a complex thermodynamic equation from first principles instead of performing simple unit analysis.

Takeaway (📌):
Dimensional analysis is your ultimate backup tool! If an admissions test presents unusual variables, convert everything to base SI units ($\text{kg, m, s}$) to check dimensional consistency.

Correct Answer: E. The container will never collide with the wall.

Fastest Approach (🚀):
Isolated system $\implies$ centre of mass remains fixed.
Container B max shift relative to CM is $0.45 \ \text{cm} < 1.0 \ \text{cm}$ distance to wall.
Container B never reaches wall C.

Matches Option E.

Step-by-Step Breakdown:

1. Centre of Mass Invariance & Isolated System Laws

• Centre of Mass Theorem: $\sum F_{\text{ext}} = 0 \implies x_{\text{CM}} = \text{constant}.$
• Spring force between A and B is internal and cannot translate centre of mass.

2. Bounded Oscillation Calculation

• Container length $L_{B} = 1.0 \ \text{cm},$ block length $L_{A} = 0.1 \ \text{cm}.$
• Internal clearance $\Delta x = 0.9 \ \text{cm}.$
• Max shift of container B: $\Delta x_{B} = \frac{m_{A}}{m_{A} + m_{B}} \Delta x = \frac{1}{2} (0.9 \ \text{cm}) = 0.45 \ \text{cm}.$

3. Distance Comparison & Option Matching

Since wall C is $1.0 \ \text{cm}$ away and $\Delta x_{B} = 0.45 \ \text{cm} < 1.0 \ \text{cm},$ container B will never collide with wall C.

Matches Option E.

Common Mistake (⚠️):
Calculating the velocity of B and assuming it travels uninterrupted toward the wall without considering that A will hit the inside of B and pull it back.

Takeaway (📌):
If a system starts at rest and experiences no external forces, its centre of mass will never move, no matter how chaotic the internal collisions are.

Correct Answer: B. 1 and 3

Fastest Approach (🚀):
Experiment 1: Identical rubber balloons $\implies$ no charge transfer $\implies$ no spark.
Experiment 3: Identical metal balloons $\implies$ conductors dissipating charge $\implies$ no spark.
Cases 1 and 3 produce no spark.

Matches Option B.

Step-by-Step Breakdown:

1. Triboelectric Charging & Material Affinity Laws

• Triboelectric Transfer: Requires two materials with different electron affinities.
• Conductors: Conductive metals allow charges to flow and dissipate, preventing localized static charge buildup.

2. Experiment-by-Experiment Analysis

• Exp 1 (Rubber on Rubber): Identical insulators $\implies$ zero net electron transfer $\implies$ NO SPARK.
• Exp 2 (Rubber on Carpet): Different insulators $\implies$ static charge accumulates $\implies$ SPARK.
• Exp 3 (Metal on Metal): Identical conductors $\implies$ charge dissipates instantly $\implies$ NO SPARK.
• Exp 4 (Rubber on Carpet, touch Metal): Charged rubber transfers charge to metal $\implies$ SPARK.

3. Scenario Selection & Option Matching

No spark in 1 and 3.

Matches Option B.

Common Mistake (⚠️):
Assuming metal rubbing on metal creates static electricity the same way insulators do.

Takeaway (📌):
Static charge builds up effectively when you rub two different insulators together.

Correct Answer: C. Its acceleration increases suddenly, then decreases until it is zero at terminal velocity

Fastest Approach (🚀):
Thruster engages $\implies$ Net upward force $= 60 + 30 - 10 = 80 \ \text{N}$ (instant acceleration spike).
As speed reverses upward, air resistance acts downward until net force $= 0.$
Acceleration increases suddenly, then decreases to zero at terminal velocity.

Matches Option C.

Step-by-Step Breakdown:

1. Dynamic Net Force & Drag Laws

• Newton's Second Law: $F_{\text{net}} = m a.$
• Drag Force Law: $\mathbf{D}$ always opposes velocity vector $\mathbf{v}.$

2. Initial Engagement & Acceleration Spike

• Weight $W = 10 \ \text{N}$ down.
• Falling drag $D_1 = 30 \ \text{N}$ up.
• Thruster $F = 60 \ \text{N}$ up.

$F_{\text{net}} = 60 + 30 - 10 = 80 \ \text{N (upward)} \implies a = 80 \ \text{m/s}^2$

3. Velocity Reversal & Terminal Velocity Stabilization

• Upward acceleration slows fall $\implies$ drag decreases.
• Once moving upward, drag acts downward: $F_{\text{net}} = 60 - 10 - D_{\text{down}} = 50 - D_{\text{down}}.$
• As upward velocity increases, $D_{\text{down}} \to 50 \ \text{N} \implies F_{\text{net}} \to 0 \implies a \to 0.$

4. Option Matching

Matches Option C.

Common Mistake (⚠️):
Forgetting that air resistance changes direction depending on which way the object is moving.

Takeaway (📌):
Air resistance always opposes the current direction of motion, meaning equations for net force dynamically flip when an object reverses direction.

Correct Answer: C. 4

Fastest Approach (🚀):
$E_{k}$ per spade $= \frac{1}{2} \times 2 \times 1^2 = 1 \ \text{J}.$
Work time $9\text{am to }1\text{pm} = 4 \ \text{hours} = 8 \times 0.5 \ \text{h}.$
Total spades $= 8 \times 1000 = 8000 \implies 8000 \ \text{J}.$
Burgers $= 8000 / 2000 = 4.$

Matches Option C.

Step-by-Step Breakdown:

1. Kinetic Energy & Work Rate Principles

• Energy per spade: $E_{k} = \frac{1}{2} m v^2.$
• Total Work: $W = N_{\text{spades}} \times E_{k}.$

2. Individual & Total Work Calculations

• $E_{k} = \frac{1}{2}(2)(1^2) = 1 \ \text{J}$ per spade.
• Work period: 9am to $1 \ \text{pm}$ $= 4$ hours.
• Rate: $1000$ spades / $0.5$ h $= 2000$ spades / h.
• Total spades $N = 2000 \times 4 = 8000$ spades $\implies W = 8000 \ \text{J}.$

3. Burger Replenishment & Option Matching

$\text{Burgers} = \frac{8000 \ \text{J}}{2000 \ \text{J/burger}} = 4\ \text{burgers}$

Matches Option C.

Common Mistake (⚠️):
Assuming the man only worked for 0.5 hours total and concluding he only needs half a burger.

Takeaway (📌):
Always double-check the total time frame given in word problems to ensure you are scaling rates correctly.

Correct Answer: B. $0.5 \ \text{W}$

Fastest Approach (🚀):
Use $P = \frac{V^2}{R}.$ Voltage is identical.
Resistance doubles ($100 \to 200\,\Omega$) $\implies$ power halves ($1 \ \text{W} \to 0.5 \ \text{W}$).

Matches Option B

Step-by-Step Breakdown:

1. Parallel Circuit Voltage Principle

In a parallel circuit, every branch experiences the exact same voltage drop ($V$) directly from the power supply:
$V_1 = V_2 = V_3 = V$

2. Selecting the Power Formula

Power dissipated in a resistor can be written as $P = I^2 R$ or $P = \frac{V^2}{R}$.
Because voltage $V$ is constant across all parallel branches, using $P = \frac{V^2}{R}$ allows us to compare power directly without needing to calculate branch currents.
Rearranging for $V^2$:
$V^2 = P_1 R_1 = P_2 R_2 = P_3 R_3$

3. Ratio & Power Calculation

Equate the power-resistance products for $R_1$ and $R_2$:
$P_1 R_1 = P_2 R_2$
Substitute $R_1 = 100 \ \Omega$, $R_2 = 200 \ \Omega$, and $P_1 = 1 \ \text{W}$:
$(1 \ \text{W}) \times (100 \ \Omega) = P_2 \times (200 \ \Omega)$
$100 = 200 P_2$
$P_2 = \frac{100}{200} = 0.5 \ \text{W}$

4. Consistency Check (confirms the given data is physically valid)

$P_1 = \frac{V^2}{R_1} = \frac{10^2}{100} = 1 \ \text{W}$, exactly matching the value given in the question.

5. Option Matching

Matches Option B.

Common Mistake (⚠️):
Using $P = I^2 R$ and assuming current is constant across branches (which is only true for series circuits), leading to the false conclusion that power doubles instead of halves.

Takeaway (📌):
For parallel circuit components: Use $P = \frac{V^2}{R}$ (power is inversely proportional to resistance). For series components: Use $P = I^2 R$ (power is directly proportional to resistance).

Correct Answer: E. $990 \ \text{J}$

Fastest Approach (🚀):
Stopping work $W = F \cdot d = 10000 \times 0.1 = 1000 \ \text{J}.$
Potential energy converted during fall $E_{p} = mgh = 0.01 \times 10 \times 100 = 10 \ \text{J}.$
Rifle energy $= 1000 - 10 = 990 \ \text{J}.$

Matches Option E.

Step-by-Step Breakdown:

1. Work-Energy & Mechanical Energy Conservation

• Work-Energy Theorem: $W_{\text{gel}} = E_{k,\text{impact}}.$
• Mechanical Energy Balance: $E_{k,\text{impact}} = E_{\text{rifle}} + E_{p}.$

2. Gel Work & Potential Energy Calculations

• Gel stopping work: $W_{\text{gel}} = 10000 \ \text{N} \times 0.1 \ \text{m} = 1000 \ \text{J}.$
• Gravitational potential energy: $E_{p} = 0.01 \ \text{kg} \times 10 \ \text{N/kg} \times 100 \ \text{m} = 10 \ \text{J}.$

3. Rifle Muzzle Energy & Option Matching

$E_{\text{rifle}} = 1000 \ \text{J} - 10\ \text{J} = 990 \ \text{J}$

Matches Option E.

Common Mistake (⚠️):
Adding the $10 \ \text{J}$ to the $1000 \ \text{J}$ instead of subtracting it, leading to $1010 \ \text{J}.$

Takeaway (📌):
Always establish a clear before-and-after energy equation to ensure you subtract or add work components correctly.

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

Fastest Approach (🚀):
Angle is given to vertical $\implies v_{y} = 20 \cos(60^\circ) = 10 \ \text{m/s}.$
Max height $h = \frac{v_{y}^2}{2g} = \frac{100}{20} = 5 \ \text{m}.$

Matches Option C.

Step-by-Step Breakdown:

1. Resolving Velocity Components

Crucial Reading Trap: The launch angle of $60^\circ$ is given to the vertical, NOT the horizontal ground.
In a right-angled triangle formed by the velocity vector:

• The vertical component $v_y$ is adjacent to the vertical angle $60^\circ$.
• Therefore, $v_y = v \cos(60^\circ)$.

$v_y = 20 \ \text{m/s} \times \cos(60^\circ) = 20 \times 0.5 = 10 \ \text{m/s}$

*(Note: Mass $m = 5 \ \text{kg}$ is extra distractor information; projectile trajectory under gravity is independent of mass.)*

2. Finding Maximum Height Using Kinematics

At the peak of its trajectory, the vertical velocity becomes zero ($v_y = 0$).
Using the SUVAT equation $v^2 = u^2 + 2as$ for vertical motion ($a = -g = -10 \ \text{m/s}^2$):
$0^2 = v_y^2 - 2gh$
$2gh = v_y^2$
$h = \frac{v_y^2}{2g}$

3. Height Calculation

Substitute $v_y = 10 \ \text{m/s}$ and $g = 10 \ \text{m/s}^2$:
$h = \frac{10^2}{2(10)} = \frac{100}{20} = 5 \ \text{m}$

4. Option Matching

Matches Option C.

Common Mistake (⚠️):
Blindly using $v \sin(60^\circ)$ because you assume angles in projectile problems are always measured from the horizontal ground, yielding $v_y = 10\sqrt{3} \implies h = 15 \ \text{m}$ (Option A).

Takeaway (📌):
Always double check the reference axis! 'Angle to the vertical' flips the standard trigonometric components: vertical becomes $v \cos\theta$ and horizontal becomes $v \sin\theta$.

Correct Answer: B. It bends away from the surface normal

Fastest Approach (🚀):
Dense (glass) to less dense (air) $\implies$ bends away from normal.

Matches Option B.

Step-by-Step Breakdown:

1. Principles of Refraction & Snell's Law

Refraction is the bending of a light wave as it passes from one medium into another with a different optical density (refractive index $n$).
Snell's Law describes this behaviour:
$n_1 \sin\theta_1 = n_2 \sin\theta_2$
where $\theta_1$ and $\theta_2$ are the angles measured relative to the surface normal (the imaginary line perpendicular to the boundary).

2. Density Comparison & Angle Behaviour

• Glass is optically denser than air ($n_{\text{glass}} \approx 1.5 > n_{\text{air}} \approx 1.0$).
• When light enters air ($n_2 < n_1$), the value of $\sin\theta_2$ must increase to satisfy Snell's Law:
• Since $\frac{n_1}{n_2} > 1$, we have $\sin\theta_2 > \sin\theta_1$, which means $\theta_2 > \theta_1$.

$\sin\theta_2 = \frac{n_1}{n_2} \sin\theta_1$

3. Direction Conclusion

Because the angle with the normal increases ($\theta_2 > \theta_1$), the light beam bends away from the surface normal as it enters the air.

4. Option Matching

Matches Option B.

Common Mistake (⚠️):
Confusing 'bends away from the normal' with 'bends away from the surface'. Bending away from the normal means leaning closer to the physical surface boundary.

Takeaway (📌):
Remember: High to Low refractive index (Dense $\to$ Rare) = Speeds up, bends away from the normal.

Correct Answer: D. $40 \ \text{W}$

Fastest Approach (🚀):
Sum resistors: $R_{\text{total}} = 10 + 20 = 30\,\Omega.$
Current $I = \frac{60}{30} = 2 \ \text{A}.$
Power $P = I^2 R_1 = 2^2 \times 10 = 40 \ \text{W}.$

Matches Option D.

Step-by-Step Breakdown:

1. Series Circuit Rules

In a series circuit:

1. The total equivalent resistance is the sum of individual resistances: $R_{\text{total}} = R_1 + R_2$.
2. The current $I$ flowing through each component is identical throughout the entire loop.

2. Calculating Total Circuit Current

• Total Resistance:
• Total Current (Ohm's Law $I = \frac{V}{R_{\text{total}}}$):

$R_{\text{total}} = 10 \ \Omega + 20 \ \Omega = 30 \ \Omega$
$I = \frac{60 \ \text{V}}{30 \ \Omega} = 2 \ \text{A}$

3. Localised Power Dissipation across $R_1$

Now use the power formula $P = I^2 R$ specifically for resistor $R_1$:
$P_1 = I^2 \cdot R_1$
$P_1 = (2 \ \text{A})^2 \times 10 \ \Omega$
$P_1 = 4 \times 10 = 40 \ \text{W}$

4. Option Matching

Matches Option D.

Common Mistake (⚠️):
Plugging the full $60 \ \text{V}$ battery voltage directly into $P = \frac{V^2}{R_1} = \frac{3600}{10} = 360 \ \text{W}$, forgetting that the $60 \ \text{V}$ is shared between both resistors.

Takeaway (📌):
In series circuits, current is constant. Always calculate total current $I = V_{\text{total}}/R_{\text{total}}$ first, then use $P = I^2 R$ for individual resistors.

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

Fastest Approach (🚀):
Series combination formula: $\frac{1}{k_{\text{eq}}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k} \implies k_{\text{eq}} = \frac{1}{2}k.$

Matches Option B.

Step-by-Step Breakdown:

1. Physics of Springs Connected in Series

When two springs are connected one after another (end-to-end in series) and a pulling force $F$ is applied to the end:

1. The force $F$ transmitted through both springs is identical ($F_1 = F_2 = F$).
2. The total extension $x_{\text{total}}$ of the combination is the sum of individual extensions ($x_{\text{total}} = x_1 + x_2$).

2. Deriving the Series Spring Formula

By Hooke's Law ($F = kx \implies x = \frac{F}{k}$):
$x_{\text{total}} = x_1 + x_2$
$\frac{F}{k_{\text{eq}}} = \frac{F}{k_1} + \frac{F}{k_2}$
Divide by $F$:
$\frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2}$

3. Calculating Equivalent Stiffness

Both springs have identical spring constants $k_1 = k_2 = k$:
$\frac{1}{k_{\text{eq}}} = \frac{1}{k} + \frac{1}{k} = \frac{2}{k}$
Taking the reciprocal of both sides:
$k_{\text{eq}} = \frac{k}{2} = \frac{1}{2} k$
Connecting springs in series halves the overall stiffness, making the system twice as easy to stretch.

4. Option Matching

Matches Option B.

Common Mistake (⚠️):
Adding the spring constants directly ($k + k = 2k$), confusing series springs with parallel springs.

Takeaway (📌):
Springs behave the EXACT OPPOSITE of electrical resistors: Series springs use reciprocal addition (weaker stiffness: $k/2$), while Parallel springs add directly (stronger stiffness: $2k$).

Correct Answer: B. Moment

Fastest Approach (🚀):
Work $= F \cdot d \implies N\cdot m.$
Moment (torque) $= F \cdot d_\perp \implies N\cdot m.$
Moment has the same dimensions as Energy.

Matches Option B.

Step-by-Step Breakdown:

1. Work & Moment Dimensional Definitions

• Work/Energy Definition: $W = F \cdot d \implies [W] = N \cdot m = kg \cdot m^2 \cdot s^{-2}.$
• Moment (Torque) Definition: $\tau = F \cdot d_\perp \implies [\tau] = N \cdot m = kg \cdot m^2 \cdot s^{-2}.$

2. Option Dimensional Auditing

• Option A (Heat capacity): $[\text{J/K}] = kg \cdot m^2 \cdot s^{-2} \cdot K^{-1}.$
• Option B (Moment): $[N \cdot m] = kg \cdot m^2 \cdot s^{-2}$ (Matches!).
• Option C (Force): $[N] = kg \cdot m \cdot s^{-2}.$
• Option D (Angular Displacement): Dimensionless.
• Option E (Moment of Inertia): $[kg \cdot m^2].$

3. Option Matching

Matches Option B.

Common Mistake (⚠️):
Assuming heat capacity has the same dimensions because it relates to heat energy, forgetting that it is energy per degree of temperature change.

Takeaway (📌):
While Energy and Moment share the same base units ($kg \cdot m^2 \cdot s^{-2}$), they represent completely different physical concepts (scalar vs vector).

Correct Answer: A. $4 \ \text{A}$

Fastest Approach (🚀):
Top branch resistance $R = 1 + (1 \parallel 1) = 1 + 0.5 = 1.5\,\Omega.$
Top branch is directly across $6 \ \text{V}$ battery $\implies I_{A} = \frac{6}{1.5} = 4 \ \text{A}.$

Matches Option A.

Step-by-Step Breakdown:

1. Parallel Branch Separation Laws

• Parallel branches connected directly across voltage $V$ operate independently.
• Branch Current Formula: $I_{\text{branch}} = \frac{V}{R_{\text{branch}}}.$

2. Top Branch Equivalent Resistance

$R_{\parallel} = \frac{1 \times 1}{1 + 1} = 0.5\,\Omega$
$R_{\text{top}} = 1\,\Omega + 0.5\,\Omega = 1.5\,\Omega$

3. Current Calculation through Point A & Option Matching

$I_{A} = \frac{6 \ \text{V}}{1.5\,\Omega} = 4 \ \text{A}$

Matches Option A.

Common Mistake (⚠️):
Trying to find the total equivalent resistance of the entire circuit first, which is unnecessary and wastes time since you only need the current in one specific branch.

Takeaway (📌):
If branches are in parallel directly across a voltage source, you can completely ignore the rest of the circuit and analyse that branch independently.

Correct Answer: C. $0.67$

Fastest Approach (🚀):
Using the formula for critical angle: $\sin(c) = \frac{n_{\text{air}}}{n_{\text{glass}}} = \frac{1.0}{1.5} = \frac{2}{3} \approx 0.67$.

Matches Option C.

Step-by-Step Breakdown:

1. Snell's Law Principle

Snell's Law states: $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$

2. Critical Angle Condition

The critical angle $c$ is the angle of incidence $\theta_1$ that produces an angle of refraction $\theta_2 = 90^{\circ}$.
Since $\sin(90^{\circ}) = 1$, the equation simplifies to:
$n_1 \sin(c) = n_2$

3. Substitution and Calculation

Given $n_1 = 1.5$ (glass) and $n_2 = 1.0$ (air):
$1.5 \sin(c) = 1.0$
$\sin(c) = \frac{1}{1.5} = \frac{2}{3} \approx 0.67$

4. Option Matching

Matches Option C.

Common Mistake (⚠️):
Calculating the ratio upside down $\frac{1.5}{1.0}$, but the sine of an angle cannot exceed $1$.

Takeaway (📌):
The sine of the critical angle is simply the ratio of the lower refractive index to the higher one.

Correct Answer: B. $6000 \ \text{N}$

Fastest Approach (🚀):
$a = \frac{20}{4} = 5 \ \text{m/s}^2$.
$F = 1200 \times 5 = 6000 \ \text{N}$.
Matches Option B.

Step-by-Step Breakdown:

1. Acceleration Calculation

The car decelerates from $20 \ \text{m/s}$ to $0 \ \text{m/s}$ in $4.0 \ \text{s}$.
$a = \frac{\Delta v}{t} = \frac{20}{4.0} = 5 \ \text{m/s}^2$

2. Force Calculation

Using Newton's Second Law:
$F = m \times a$
$F = 1200 \ \text{kg} \times 5 \ \text{m/s}^2 = 6000 \ \text{N}$

3. Option Matching

Matches Option B.

Common Mistake (⚠️):
Squaring the velocity or forgetting to divide by the time.

Takeaway (📌):
Always write down known variables ($u, v, t, m$) and link them through standard equations.

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

Fastest Approach (🚀):
Left tension $T_{L} = \frac{mg}{2},$ Right tension $T_{R} = \frac{3mg}{2}.$
Upward spring force $= T_{L} + T_{R} = 2\ \text{mg} = 40 \ \text{N}.$
Spring extension $x = \frac{40}{200} = 0.2 \ \text{m} = 20 \ \text{cm}.$

Matches Option C.

Step-by-Step Breakdown:

1. Movable Pulley & Static Equilibrium Laws

• Movable Pulley Mechanical Advantage: $T = \frac{M_{\text{load}} g}{2}.$
• Hooke's Law: $F_{\text{spring}} = k x.$

2. Rope Tension Analysis

• Left rope tension ($m = 2 \ \text{kg}$): $T_{L} = \frac{mg}{2}.$
• Right rope tension ($3m = 6 \ \text{kg}$): $T_{R} = \frac{3mg}{2}.$
• Central spring upward pull:

$F_{\text{spring}} = T_{L} + T_{R} = \frac{mg}{2} + \frac{3mg}{2} = 2\ \text{mg}$

3. Spring Extension Calculation & Option Matching

Substitute parameters ($m = 2 \ \text{kg}, g = 10 \ \text{m/s}^2, k = 200 \ \text{N/m}$):
$F_{\text{spring}} = 2(2)(10) = 40 \ \text{N}$
$kx = 40 \implies 200x = 40 \implies x = 0.2 \ \text{m} = 20 \ \text{cm}$

Matches Option C.

Common Mistake (⚠️):
Assuming the tension in the ropes is equal to the full weight of the masses ($mg$ and $3\ \text{mg}$), forgetting that movable pulleys halve the required tension.

Takeaway (📌):
Always trace a rope's path carefully around movable pulleys to verify how many rope segments are supporting the load.

Correct Answer: C. $4.0\times10^{14}$

Fastest Approach (🚀):
$n = \frac{380}{9.61} \approx 39.54.$
Initial amount $N_0 = 500 \times 2^{39.54} \approx 500 \times 7.99\times10^{11} \approx 4.0 \times 10^{14}.$

Matches Option C.

Step-by-Step Breakdown:

1. Exponential Radioactive Decay Laws

• Reverse Decay Formula: $N_0 = N(t) \cdot 2^{n}.$
• Half-life count: $n = \frac{t}{t_{1/2}}.$

2. Half-Life Count & Reverse Decay Setup

$n = \frac{380}{9.61} \approx 39.54$
$N_0 = 500 \times 2^{39.54}$

3. Logarithmic & Power Estimation & Option Matching

Using $\log_{10}(2) \approx 0.301$: $2^{39.54} \approx 10^{39.54 \times 0.301} = 10^{11.90} \approx 7.99\times10^{11}.$
$N_0 \approx 500 \times 7.99\times10^{11} \approx 4.0 \times 10^{14}$

Matches Option C.

Common Mistake (⚠️):
Using division instead of multiplication and solving for how many atoms are left in the future instead of how many there were in the past.

Takeaway (📌):
Memorize the approximation $2^{10} \approx 10^3$; it is invaluable for mental math involving large exponents.

Correct Answer: A. $2.0 \ \Omega, \ 8.0 \ \text{W}$

Fastest Approach (🚀):
$R_T = \frac{12}{2.0} = 6.0 \ \Omega$.
Lamp $R = 6.0 - 4.0 = 2.0 \ \Omega$.
Lamp Power $P = I^2 R = (2.0)^2 \times 2.0 = 4.0 \times 2.0 = 8.0 \ \text{W}$.
Matches Option A.

Step-by-Step Breakdown:

1. Total Resistance

Using Ohm's Law for the whole circuit:
$R_T = \frac{V}{I} = \frac{12 \ \text{V}}{2.0 \ \text{A}} = 6.0 \ \Omega$

2. Lamp Resistance

In a series circuit, $R_T = R_1 + R_2$:
$6.0 \ \Omega = 4.0 \ \Omega + R_{\text{lamp}}$
$R_{\text{lamp}} = 2.0 \ \Omega$

3. Lamp Power Dissipation

Using the power equation $P = I^2 R$:
$P = (2.0 \ \text{A})^2 \times 2.0 \ \Omega = 4.0 \times 2.0 = 8.0 \ \text{W}$

4. Option Matching

Matches Option A.

Common Mistake (⚠️):
Calculating the total power of the circuit ($P = VI = 24 \ \text{W}$) instead of the power dissipated by the lamp alone.

Takeaway (📌):
When calculating power for a specific component, always use the resistance or voltage specific to that component.

Correct Answer: C. $1 \ \text{cm}$

Fastest Approach (🚀):
Displaced water volume $V = \frac{300}{1000} = 0.3 \ \text{m}^3.$
Base area $A = 10 \times 3 = 30 \ \text{m}^2.$
Submerged depth $d = \frac{0.3}{30} = 0.01 \ \text{m} = 1 \ \text{cm}.$

Matches Option C.

Step-by-Step Breakdown:

1. Archimedes' Principle & Buoyancy Equilibrium

Displaced water mass $m_{w} = \text{Boat mass } m_{b} = 300 \ \text{kg}.$

2. Displaced Volume Calculation

$V_{\text{displaced}} = \frac{m_{b}}{\rho_{w}} = \frac{300}{1000} = 0.3 \ \text{m}^3$

3. Submerged Depth Calculation & Option Matching

Base area $A = 10 \times 3 = 30 \ \text{m}^2.$
$d = \frac{V_{\text{displaced}}}{A} = \frac{0.3}{30} = 0.01\ \text{m} = 1 \ \text{cm}$

Matches Option C.

Common Mistake (⚠️):
Using the $50 \ \text{cm}$ wall height in the calculation. This is extraneous information meant to trick you.

Takeaway (📌):
Volume of water displaced = Mass of floating object / Density of water.

Correct Answer: B. $0.02 \ \text{s}$

Fastest Approach (🚀):
Exit speed $v = 1.25 \times 40 = 50 \ \text{m/s}.$
Direction change $\implies \Delta v = 50 - (-40) = 90 \ \text{m/s}.$
$F \Delta t = m \Delta v \implies 270 \Delta t = 0.06 \times 90 = 5.4 \implies \Delta t = 0.02 \ \text{s}.$

Matches Option B.

Step-by-Step Breakdown:

1. Impulse-Momentum Theorem & Vector Velocity

• Impulse Formula: $F \cdot \Delta t = m \Delta v.$
• Reversal $\implies \Delta v = v_{\text{exit}} - (-v_{\text{entry}}) = v_{\text{exit}} + v_{\text{entry}}.$

2. Parameter Calculation

• Mass $m = 0.06 \ \text{kg}.$
• Exit speed $v = 40 \times 1.25 = 50 \ \text{m/s}.$
• Velocity change $\Delta v = 50 - (-40) = 90 \ \text{m/s}.$

3. Contact Time Calculation & Option Matching

$270 \times \Delta t = 0.06 \times 90 = 5.4 \implies \Delta t = \frac{5.4}{270} = 0.02 \ \text{s}$

Matches Option B.

Common Mistake (⚠️):
Subtracting the speeds ($50 - 40 = 10$) instead of adding them when calculating $\Delta v$ for an object bouncing back.

Takeaway (📌):
Whenever an object bounces or reverses direction, the change in velocity is the sum of the initial and final speeds.

Correct Answer: E. It doesn't

Fastest Approach (🚀):
Light travels from $n=1.45$ to $n=1.66$ (lower to higher index).
TIR is physically impossible $\implies$ "It doesn't".

Matches Option E.

Step-by-Step Breakdown:

1. Total Internal Reflection Prerequisites

TIR occurs ONLY when light travels from optically denser ($n_1$) to less dense ($n_2$) medium ($n_1 > n_2$).

2. Refractive Index Comparison & Conclusion

$n_1 = 1.45 < n_2 = 1.66 \implies$ Light refracts into glucose solution for all incident angles.
Total internal reflection cannot occur.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. 29 degrees — Calculation Error
  This angle is calculated if you mistakenly take the arcsin of (1 - 1.45/1.66) or similar incorrect math.
• B. 42 degrees — Memorization Error
  This angle represents the critical angle of glass (n=1.5) to air (n=1), a common memorized value that doesn't apply here.
• C. 61 degrees — Misapplied Formula
  Calculated by blindly using arcsin(1.45/1.66) = 60.8 degrees. This is the critical angle if light was going from glucose TO glass.
• D. 85 degrees — Mathematical Error
  An incorrect angle likely derived from taking arcsin(1.66/1.45), which is mathematically impossible and might lead a student to guess a large angle.

Common Mistake (⚠️):
Plugging the numbers blindly into the critical angle formula ($\sin(\theta_c) = n_{\text{low}}/n_{\text{high}}$) to get $\arcsin(1.45/1.66) \approx 61^\circ$ (Option C), without checking if TIR is physically possible.

Takeaway (📌):
Always verify the direction of the light ray before calculating TIR; it only works High-to-Low.

Correct Answer: A. $\frac{4}{3} \ \Omega$

Fastest Approach (🚀):
Original resistance $= 12 \ \Omega$.
Resistance of each third $= \frac{12}{3} = 4 \ \Omega$.
Three $4 \ \Omega$ resistors in parallel: $R_{\text{eq}} = \frac{4}{3} \ \Omega$.

Matches Option A.

Step-by-Step Breakdown:

1. Resistance and Length Proportionality

The resistance $R$ of a uniform wire is directly proportional to its length ($R \propto L$).
If a $12 \ \Omega$ wire is cut into three equal pieces, each piece has one-third the original length.
Resistance of one piece: $R_1 = \frac{12}{3} = 4 \ \Omega$.

2. Parallel Circuit Laws

For $n$ identical resistors of resistance $R$ connected in parallel, the equivalent resistance is:
$R_{\text{eq}} = \frac{R}{n}$

3. Equivalent Resistance Calculation

We have three resistors ($n=3$), each with resistance $R_1 = 4 \ \Omega$.
$R_{\text{eq}} = \frac{4}{3} \ \Omega$

4. Option Matching

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $3 \ \Omega$ — Calculation Error
  Derived by mistakenly dividing the original $12 \ \Omega$ by 4 instead of 3 (or doing $12/(3+1)$).
• C. $4 \ \Omega$ — Incomplete Problem Solving
  Calculated by correctly identifying that each piece has $4 \ \Omega$ of resistance, but forgetting to then connect them in parallel. $4 \ \Omega$ is the resistance of just one piece.
• D. $12 \ \Omega$ — Reading Comprehension
  This is the original resistance. If a student completely misinterprets 'equivalent resistance' and just answers with the original value.
• E. $36 \ \Omega$ — Conceptual Error
  Calculated by multiplying the original resistance by 3, which is the equivalent resistance if you had three identical $12 \ \Omega$ wires connected in series.

Common Mistake (⚠️):
Adding the resistors in series to get $12 \ \Omega$, or simply taking $12/3 = 4 \ \Omega$ and forgetting they are then placed in parallel.

Takeaway (📌):
When a uniform resistor $R$ is cut into $n$ pieces and connected in parallel, the new equivalent resistance is always $\frac{R}{n^2}$.