Olympiad & IIT-JEE Advanced Problem Sets: Master Solutions

How to Use This Problem Set

Each problem in this curated collection is designed to test deep physical intuition, coordinate choices, and rigorous mathematical execution.
Recommended Workflow: Attempt each problem on paper for 15–20 minutes before opening the "Click to view complete step-by-step solution" dropdown.

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Volume 1: Advanced Mechanics

Problem 1.1: The Rolling Spool Paradox on an Incline

INPhO / IPhO Masterclass

Problem Statement:
A spool of total mass $M$, outer radius $R$, and inner cylinder radius $r$ ($r<R$) has a radius of gyration $k$ about its central axis ($I_{\text{cm}}=M{k}^2$). It rests on a rough plane inclined at an angle $\theta$ to the horizontal. A light inextensible thread wound around the inner axle is pulled with a constant tension $T$ at an angle $\varphi$ relative to the inclined plane as shown. The coefficient of static friction is sufficiently large to prevent slipping.

Multi-Mode DiagramProblem 1.1: Rolling Spool Dynamics (Multi-Mode)

Option 1: Publication-Grade Scientific Vector SVG

Free-body force decomposition on incline showing normal force N, static friction fs, tension T, and Instantaneous Axis of Rotation C.

Linear Acceleration: $a_{\text{cm}} = \frac{g\sin\theta}{1 + I_{\text{cm}}/(MR^2)} = \frac{2}{3}g\sin\theta$

Critical Tension Angle: $\phi_c = \arccos(r/R)$

Option 2: Interactive 60 FPS Canvas Physics Engine

Real-time 60 FPS pure rolling simulation on incline.

Incline Angle θ: 25 °

Acceleration ($a_{\text{cm}}$):2.76 m/s²

Option 3: Interactive 3D Spatial Viewport

3D rotatable gyroscope disc.

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

Precession Equation:$\vec{\tau} = \frac{d\vec{L}}{dt}$

Option 5: Reactive Parametric Physics with Live Calculus

Parametric angle sliders with live torque calculation.

Incline Angle θ: 25 °

Pull Angle φ: 45 °

Critical Angle ($\phi_c$):$62.5°$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

1. Derive the acceleration $a_{\text{cm}}$ of the spool down the incline.
2. Find the critical tension $T_0$ for which the spool remains in static equilibrium ($a_{\text{cm}}=0$).
3. Find the critical angle ${\varphi }_c$ at which pulling the thread causes the spool to slide rather than roll.

💡 Interactive Clue SystemClues for Problem 1.1: The Rolling Spool Paradox

Try solving with Hint 1 before revealing Hint 2 or 3!

🟡 Hint 1: Physical Intuition & Coordinate Choice

Don't set up complicated simultaneous equations at the Center of Mass with unknown friction $f_s$ and normal force $N$. Shift your reference frame to the contact point $C$ on the incline (IAOR)!

🟠 Hint 2: Key Governing Equations

Compute perpendicular distance from $C$ to the line of action of tension $T$: $d_\\perp = R\\cos\\phi - r$. The net torque about $C$ is $\\tau_C = M g R \\sin\\theta - T(R\\cos\\phi - r) = I_C \\alpha$.

🟢 Hint 3: Full Step-by-Step Derivation

Equilibrium ($a_{\\text{cm}} = 0$) occurs when net torque vanishes: $T_0(R\\cos\\phi - r) = M g R \\sin\\theta \\implies T_0 = \\frac{M g R \\sin\\theta}{R\\cos\\phi - r}$. Pure sliding occurs when $d_\\perp = 0 \\implies \\cos\\phi_c = r/R$.

💡 Click to view complete step-by-step solution

Step 1: Choose the Reference Point (Instantaneous Axis of Rotation - IAOR)

The instantaneous point of contact $C$ with the incline has zero velocity ($v_C=0$). Taking torques directly about $C$ automatically eliminates the unknown normal force $N$ and static friction force $f$!

Step 2: Write the Equation of Motion about Contact Point $C$

1. Moment of Inertia about $C$ (Parallel Axis Theorem):

   $$I_C=I_{\text{cm}}+M{R}^2=M(k^2+R^2)$$

2. Torque of Gravity about $C$: Gravity acts at the Center of Mass (distance $R$ perpendicularly from $C$ along the incline):

   $${\tau }_{\text{gravity},C}=(Mg\sin \theta )\cdot R\text{(Acts down the incline, promoting clockwise roll)}$$

3. Torque of Tension $T$ about $C$: The position vector from $C$ to the center $O$ is ${\overrightarrow{r}}_{O/C}=R{\hat{j}}^{\prime }$ (normal to incline). The line of action of tension $T$ passes at perpendicular distance from $C$:

   $$d_{\perp }=R\cos \varphi -r$$$${\tau }_{\text{tension},C}=T(R\cos \varphi -r)\text{(Acts up the incline if }R\cos \varphi >r\text{)}$$

4. Net Torque about $C$:

   $${\tau }_{\text{net},C}=MgR\sin \theta -T(R\cos \varphi -r)$$

5. Angular Acceleration $\alpha$:

   $$I_C\alpha ={\tau }_{\text{net},C}⟹M(k^2+R^2)\alpha =MgR\sin \theta -T(R\cos \varphi -r)$$$$\alpha =\frac{MgR\sin \theta -T(R\cos \varphi -r)}{M(k^2+R^2)}$$

6. Linear Acceleration of Center of Mass $a_{\text{cm}}=\alpha R$:

   $$\boxed{{\displaystyle a_{\text{cm}}=\frac{g\sin \theta -\frac{T}{M}(\cos \varphi -\frac{r}{R})}{1+\frac{k^2}{R^2}}}}$$

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Step 3: Condition for Static Equilibrium ($a_{\text{cm}}=0$)

Setting the numerator to zero:

$$Mg\sin \theta -T_0(\cos \varphi -\frac{r}{R})=0⟹\boxed{{\displaystyle T_0=\frac{Mg\sin \theta }{\cos \varphi -\frac{r}{R}}}}$$

Step 4: The Critical Angle ${\varphi }_c$

If $\cos \varphi =\frac{r}{R}$, the line of action of tension passes directly through the contact point $C$!

$${\tau }_{\text{tension},C}=0⟹\boxed{{\displaystyle \cos {\varphi }_c=\frac{r}{R}⟹{\varphi }_c=\arccos \left(\frac{r}{R}\right)}}$$

At this angle, tension creates zero torque about the contact point, so tension cannot cause or prevent rotation!

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Problem 1.2: Separation Angle on a Movable Hemisphere

JEE Advanced Multi-Concept

Problem Statement:
A small particle of mass $m$ is placed at the apex of a smooth solid hemisphere of mass $M$ and radius $R$. The hemisphere rests on a frictionless horizontal floor. The particle is given an infinitesimal horizontal nudge. Determine the angle $\theta$ from the vertical at which the particle loses contact with the hemisphere ($N=0$).

<!-- Hemisphere (Radius R, Mass M) --> <path d="M 110 150 A 100 100 0 0 1 310 150 Z" fill="url(#spoolGrad)" stroke="var(--vp-c-brand-1)" stroke-width="2.5"/> <circle cx="210" cy="150" r="3" fill="var(--vp-c-brand-1)"/> <text x="205" y="130" fill="var(--vp-c-text-1)" font-size="13" font-weight="700">M</text> <!-- Particle m at angle theta from apex --> <g transform="translate(210, 150) rotate(35)"> <line x1="0" y1="0" x2="0" y2="-100" stroke="var(--vp-c-text-3)" stroke-dasharray="3 3"/> <circle cx="0" cy="-100" r="10" fill="#ef4444" stroke="var(--vp-c-text-1)" stroke-width="1.5"/> <text x="-5" y="-96" fill="#fff" font-size="10" font-weight="700">m</text> </g> <!-- Angle theta arc --> <path d="M 210 70 A 80 80 0 0 1 255 78" fill="none" stroke="#f59e0b" stroke-width="2"/> <text x="228" y="65" fill="#f59e0b" font-size="12" font-weight="700">&theta;</text> 

Figure Movable hemisphere on frictionless floor: Particle slides down until coupled normal reaction N(theta) drops to zero.

💡 Click to view complete step-by-step solution

Step 1: Set up the Coordinates and Constraints

Let the horizontal displacement of the hemisphere be $X(t)$ (moving leftward, $-x$).
In polar coordinates relative to the hemisphere center:

• Particle position relative to hemisphere: $x_r=R\sin \theta ,y_r=R\cos \theta$.
• Absolute horizontal position of particle: $x=R\sin \theta -X$.
• Absolute vertical position of particle: $y=R\cos \theta$.

Step 2: Conservation of Horizontal Momentum

Since no external horizontal force acts on the (particle + hemisphere) system:

$$P_x=m\dot{x}-M\dot{X}=0⟹m(R\dot{\theta }\cos \theta -\dot{X})-M\dot{X}=0$$$$\dot{X}=\frac{mR\cos \theta }{M+m}\dot{\theta }$$

The horizontal velocity of the particle is:

$$\dot{x}=R\dot{\theta }\cos \theta -\dot{X}=R\dot{\theta }\cos \theta (1-\frac{m}{M+m})=\left(\frac{M}{M+m}\right)R\dot{\theta }\cos \theta$$

Step 3: Conservation of Mechanical Energy

$$\mathrm{\Delta }U=mgR(1-\cos \theta )$$$$\mathrm{\Delta }K=\frac{1}{2}M{\dot{X}}^2+\frac{1}{2}m({\dot{x}}^2+{\dot{y}}^2)$$

Substituting $\dot{y}=-R\dot{\theta }\sin \theta$ and $\dot{X}$:

$$mgR(1-\cos \theta )=\frac{1}{2}m{R}^2{\dot{\theta }}^2({\sin }^2\theta +\frac{M{\cos }^2\theta }{M+m})$$$${\dot{\theta }}^2=\frac{2g(1-\cos \theta )}{R({\sin }^2\theta +\frac{M}{M+m}{\cos }^2\theta )}=\frac{2g(1-\cos \theta )}{R(1-\frac{m}{M+m}{\cos }^2\theta )}$$

Step 4: Normal Force Separation Condition ($N=0$)

In the accelerated frame of the hemisphere, the radial equation of motion for $m$ at the moment of detachment ($N=0$) gives:

$$mg\cos \theta =mR{\dot{\theta }}^2+m\ddot{X}\sin \theta$$

Solving the coupled equations at $N=0$ yields the cubic polynomial:

$$\boxed{{\displaystyle \frac{m}{M+m}{\cos }^3\theta -3\cos \theta +2=0}}$$

• Limiting Case 1 (Fixed Hemisphere $M\to \mathrm{\infty }$):$$\frac{m}{\mathrm{\infty }}{\cos }^3\theta -3\cos \theta +2=0⟹-3\cos \theta +2=0⟹\cos \theta =\frac{2}{3}⟹\theta \approx {48.2}^{\circ }$$
• Limiting Case 2 (Equal masses $M=m$):$$\frac{1}{2}{\cos }^3\theta -3\cos \theta +2=0⟹\cos \theta \approx 0.701⟹\theta \approx {45.5}^{\circ }$$

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Volume 2: Classical Electrodynamics

Problem 2.1: Superconducting Ring in a Quadrupole Field

International Physics Olympiad (IPhO Tier)

Problem Statement:
A thin circular superconducting ring of radius $a$, mass $m$, and self-inductance $L$ is placed in a magnetic quadrupole field given by:

$$\overrightarrow{B}(x,y,z)=\alpha (x\hat{i}-y\hat{j})$$

The ring lies in the $xy$-plane, centered at $(x_0,0,0)$ with its normal initially along $\hat{k}$. If the ring is displaced slightly along the $x$-axis and released from rest, prove that it executes Simple Harmonic Motion (SHM) and find its angular frequency $\omega$.

<line x1="50" y1="120" x2="50" y2="20" stroke="var(--vp-c-text-1)" stroke-width="2" marker-end="url(#arrow-blue)"/> <text x="30" y="25" fill="var(--vp-c-text-1)" font-size="12" font-weight="700">y</text> <!-- Superconducting Ring (centered at x_0) --> <ellipse cx="230" cy="120" rx="40" ry="20" fill="none" stroke="var(--vp-c-brand-1)" stroke-width="3"/> <circle cx="230" cy="120" r="3" fill="#ef4444"/> <text x="220" y="150" fill="var(--vp-c-text-2)" font-size="12" font-weight="700">x_0</text> <text x="200" y="90" fill="var(--vp-c-brand-1)" font-size="12" font-weight="700">Ring (m, L)</text> <!-- Quadrupole Field Vectors --> <line x1="120" y1="60" x2="160" y2="60" stroke="#3b82f6" stroke-width="2" marker-end="url(#arrow-blue)"/> <line x1="300" y1="60" x2="350" y2="60" stroke="#3b82f6" stroke-width="2" marker-end="url(#arrow-blue)"/> <text x="180" y="45" fill="#3b82f6" font-size="12" font-weight="700">B = &alpha;(x i - y j)</text> 

Figure Superconducting ring in quadrupole field B = alpha (x i - y j): Flux freezing induces restoring current I(x) causing harmonic oscillation.

💡 Click to view complete step-by-step solution

Step 1: Flux Freezing Theorem in Superconductors

In an ideal superconductor, electrical resistance is strictly zero ($R=0$). By Faraday's Law:

$$\mathcal{E}=-\frac{d{\mathrm{\Phi }}_{\text{total}}}{dt}=IR=0⟹{\mathrm{\Phi }}_{\text{total}}=\text{constant}={\mathrm{\Phi }}_0$$

The total magnetic flux through the ring is:

$${\mathrm{\Phi }}_{\text{total}}={\mathrm{\Phi }}_{\text{ext}}(x)+LI(x)={\mathrm{\Phi }}_0$$

If the ring starts with zero current at initial position $x_0$ where ${\mathrm{\Phi }}_{\text{ext}}(x_0)=0$, then ${\mathrm{\Phi }}_0=0$, so:

$$I(x)=-\frac{{\mathrm{\Phi }}_{\text{ext}}(x)}{L}$$

Step 2: Compute External Flux ${\mathrm{\Phi }}_{\text{ext}}(x)$

Since $\overrightarrow{B}=\alpha x\hat{i}-\alpha y\hat{j}$, the field in the $xy$-plane has zero $z$-component ($B_z=0$).
However, by $\mathrm{\nabla }\cdot \overrightarrow{B}=0$:

$$\frac{\partial B_x}{\partial x}+\frac{\partial B_y}{\partial y}+\frac{\partial B_z}{\partial z}=\alpha -\alpha +\frac{\partial B_z}{\partial z}=0⟹\frac{\partial B_z}{\partial z}=0$$

When the ring tilts or translates across the gradient:

$${\mathrm{\Phi }}_{\text{ext}}(x)=\pi a^2(\alpha x)$$$$I(x)=-\frac{\pi a^2\alpha }{L}x$$

Step 3: Magnetic Force on the Ring

The magnetic dipole moment of the ring is:

$$\overrightarrow{M}(x)=I(x)(\pi a^2)\hat{k}=-\frac{(\pi a^2{)}^2\alpha }{L}x\hat{k}$$

The net force acting on the magnetic dipole in a non-uniform field is:

$$\overrightarrow{F}=\mathrm{\nabla }(\overrightarrow{M}\cdot \overrightarrow{B})=-\frac{(\pi a^2{)}^2{\alpha }^2}{L}x\hat{i}$$

Step 4: Equation of Motion & SHM Frequency

$$m\frac{d^{2}x}{d{t}^2}=-\left[\frac{{\pi }^2{a}^4{\alpha }^2}{L}\right]x$$$$\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$$$$\boxed{{\displaystyle \omega =\sqrt{\frac{{\pi }^2{a}^4{\alpha }^2}{mL}}=\frac{\pi a^2\alpha }{\sqrt{mL}}}}$$

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Problem 2.2: Non-Uniform Dielectric Sphere in a Uniform Field

JEE Advanced Theoretical Drill

Problem Statement:
A solid dielectric sphere of radius $R$ is placed in an initially uniform electric field ${\overrightarrow{E}}_0=E_0\hat{k}$. The relative permittivity inside the sphere varies radially as:

$${\epsilon }_r(r)=1+\beta \frac{r}{R}$$

1. Find the polarization vector $\overrightarrow{P}(\overrightarrow{r})$ inside the sphere.
2. Determine the bound volume charge density ${\rho }_b(r)$ and bound surface charge density ${\sigma }_b(\theta )$.

💡 Click to view complete step-by-step solution

Step 1: Electric Displacement & Polarization

The electric displacement is related to electric field by:

$$\overrightarrow{D}={\epsilon }_0{\epsilon }_r(r)\overrightarrow{E}$$

The polarization field $\overrightarrow{P}$ is:

$$\overrightarrow{P}=\overrightarrow{D}-{\epsilon }_0\overrightarrow{E}={\epsilon }_0({\epsilon }_r(r)-1)\overrightarrow{E}={\epsilon }_0\left(\beta \frac{r}{R}\right)\overrightarrow{E}$$

Step 2: Internal Electric Field

To first order in dielectric perturbation, the uniform field inside the sphere is:

$${\overrightarrow{E}}_{\text{in}}=E_{\text{in}}\hat{k}$$

Thus the polarization vector is:

$$\boxed{{\displaystyle \overrightarrow{P}(r,\theta )={\epsilon }_0\beta \frac{r}{R}{E}_{\text{in}}\hat{k}={\epsilon }_0\beta \frac{r}{R}{E}_{\text{in}}(\cos \theta {\hat{e}}_r-\sin \theta {\hat{e}}_{\theta })}}$$

Step 3: Bound Volume Charge Density ${\rho }_b$

$${\rho }_b=-\mathrm{\nabla }\cdot \overrightarrow{P}$$

In spherical coordinates:

$$\mathrm{\nabla }\cdot \overrightarrow{P}=\frac{1}{r^2}\frac{\partial }{\partial r}(r^2{P}_r)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }(\sin \theta P_{\theta })$$$$P_r={\epsilon }_0\beta \frac{r}{R}{E}_{\text{in}}\cos \theta ,P_{\theta }=-{\epsilon }_0\beta \frac{r}{R}{E}_{\text{in}}\sin \theta$$$$\frac{1}{r^2}\frac{\partial }{\partial r}\left(\frac{{\epsilon }_0\beta E_{\text{in}}\cos \theta }{R}{r}^3\right)=\frac{3{\epsilon }_0\beta E_{\text{in}}\cos \theta }{R}$$$$\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }(-\frac{{\epsilon }_0\beta r{E}_{\text{in}}}{R}{\sin }^2\theta )=-\frac{2{\epsilon }_0\beta E_{\text{in}}\cos \theta }{R}$$$$\mathrm{\nabla }\cdot \overrightarrow{P}=\frac{3{\epsilon }_0\beta E_{\text{in}}\cos \theta }{R}-\frac{2{\epsilon }_0\beta E_{\text{in}}\cos \theta }{R}=\frac{{\epsilon }_0\beta E_{\text{in}}\cos \theta }{R}$$$$\boxed{{\displaystyle {\rho }_b(r,\theta )=-\frac{{\epsilon }_0\beta E_{\text{in}}\cos \theta }{R}}}$$

Step 4: Bound Surface Charge Density ${\sigma }_b$

$${\sigma }_b=\overrightarrow{P}(R)\cdot \hat{n}=P_r(R)=\boxed{{\displaystyle {\epsilon }_0\beta E_{\text{in}}\cos \theta }}$$

(Notice: ${\oint }_{\text{surface}}{\sigma }_{b}dA+{\int }_{\text{volume}}{\rho }_{b}dV=0$, guaranteeing net bound charge is strictly zero!)

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Volume 3: Waves, Optics & Thermodynamics

Problem 3.1: Sound Ray Trajectory in a Thermal Gradient Atmosphere

National Physics Olympiad (INPhO Level)

Problem Statement:
In the lower atmosphere, the absolute temperature decreases linearly with altitude $y$ as:

$$T(y)=T_0(1-\alpha y)$$

A sound source located on the ground at $(0,0)$ emits a sound ray at an initial angle ${\theta }_0$ relative to the vertical.

1. Find the trajectory equation $y(x)$ of the sound ray.
2. Determine the maximum altitude $y_{max}$ reached by the sound ray before it refracts back toward Earth.

<!-- Altitude Axis --> <line x1="60" y1="150" x2="60" y2="20" stroke="var(--vp-c-text-1)" stroke-width="2" marker-end="url(#arrow-blue)"/> <text x="30" y="25" fill="var(--vp-c-text-1)" font-size="12" font-weight="700">y</text> <!-- Parabolic Refracting Ray --> <path d="M 60 150 Q 180 30 320 150" fill="none" stroke="#f59e0b" stroke-width="3"/> <!-- Launch angle indicator --> <line x1="60" y1="150" x2="110" y2="90" stroke="#f59e0b" stroke-width="2" marker-end="url(#arrow-red)"/> <text x="75" y="115" fill="var(--vp-c-brand-1)" font-size="12" font-weight="700">&theta;_0</text> <!-- Apex Point y_max --> <circle cx="190" cy="90" r="5" fill="#ef4444"/> <line x1="60" y1="90" x2="190" y2="90" stroke="var(--vp-c-text-3)" stroke-dasharray="3 3" stroke-width="1.2"/> <text x="140" y="80" fill="#ef4444" font-size="12" font-weight="700">Apex (y_max)</text> 

Figure Acoustic ray path in stratified atmosphere: Continuous Snell's law bends the ray into a parabolic trajectory reaching apex y_max = cos^2(theta_0) / alpha.

💡 Click to view complete step-by-step solution

Step 1: Acoustic Speed Variation with Altitude

The speed of sound in air is $v(T)=\sqrt{\frac{\gamma RT}{M}}$:

$$v(y)=v_0\sqrt{1-\alpha y}(\text{where }{v}_0=\sqrt{\frac{\gamma R{T}_0}{M}})$$

Step 2: Continuous Snell's Law for Acoustic Rays

For any stratified medium where speed depends on altitude $y$:

$$\frac{\sin \theta (y)}{v(y)}=\text{constant}=\frac{\sin {\theta }_0}{v_0}$$$$\sin \theta (y)=\left(\frac{v(y)}{v_0}\right)\sin {\theta }_0=\sqrt{1-\alpha y}\sin {\theta }_0$$

Step 3: Maximum Altitude $y_{max}$

At the apex of the ray, the trajectory becomes purely horizontal ($\theta ={90}^{\circ }⟹\sin \theta =1$):

$$1=\sqrt{1-\alpha y_{max}}\sin {\theta }_0⟹1-\alpha y_{max}=\frac{1}{{\sin }^2{\theta }_0}$$$$\alpha y_{max}=1-{\sin }^2{\theta }_0={\cos }^2{\theta }_0⟹\boxed{{\displaystyle y_{max}=\frac{{\cos }^2{\theta }_0}{\alpha }}}$$

Step 4: Differential Equation of the Ray Path

From geometry: $\tan \theta =\frac{dx}{dy}$:

$$\frac{dx}{dy}=\frac{\sin \theta }{\cos \theta }=\frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }}=\frac{\sqrt{1-\alpha y}\sin {\theta }_0}{\sqrt{1-(1-\alpha y){\sin }^2{\theta }_0}}=\frac{\sqrt{1-\alpha y}\sin {\theta }_0}{\sqrt{{\cos }^2{\theta }_0+\alpha y{\sin }^2{\theta }_0}}$$

Integrating gives a parabolic acoustic trajectory:

$$\boxed{{\displaystyle x(y)=\frac{2}{\alpha \sin {\theta }_0}[\cos {\theta }_0-\sqrt{{\cos }^2{\theta }_0-\alpha y{\sin }^2{\theta }_0}]}}$$

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Problem 3.2: Non-Ideal Gas Cycle Efficiency Optimization

JEE Advanced Multi-Concept

Problem Statement:
One mole of a real gas obeying the equation of state $P(V-b)=RT$ with constant molar heat capacity $C_v$ undergoes a four-stroke cycle:

• $1\to 2$: Isothermal expansion at $T_H$ from $V_1$ to $V_2$.
• $2\to 3$: Adiabatic expansion from $T_H$ to $T_C$.
• $3\to 4$: Isothermal compression at $T_C$ from $V_3$ to $V_4$.
• $4\to 1$: Adiabatic compression from $T_C$ to $T_H$.

Prove that the thermal efficiency $\eta$ of this cycle is identical to an ideal Carnot efficiency: $\eta =1-\frac{T_C}{T_H}$.

💡 Click to view complete step-by-step solution

Step 1: Work Done along Isothermal Paths

For the isothermal expansion $1\to 2$ at $T_H$:

$$P=\frac{R{T}_H}{V-b}⟹W_{12}={\int }_{V_1}^{V_2}\frac{R{T}_H}{V-b}dV=R{T}_H\ln \left(\frac{V_2-b}{V_1-b}\right)$$

Since $T=\text{const}⟹\mathrm{\Delta }U=0$:

$$Q_{\text{in}}=Q_{12}=W_{12}=R{T}_H\ln \left(\frac{V_2-b}{V_1-b}\right)$$

For the isothermal compression $3\to 4$ at $T_C$:

$$Q_{\text{out}}=|Q_{34}|=R{T}_C\ln \left(\frac{V_3-b}{V_4-b}\right)$$

Step 2: Adiabatic Paths Relation

Along an adiabatic path for $P(V-b)=RT$:

$$dQ=C_{v}dT+PdV=0⟹C_{v}dT+\frac{RT}{V-b}dV=0⟹\frac{dT}{T}+\frac{R}{C_v}\frac{dV}{V-b}=0$$

Integrating yields:

$$T(V-b{)}^{R/C_v}=T(V-b{)}^{\gamma -1}=\text{constant}$$

Applying this to paths $2\to 3$ and $4\to 1$:

$$T_H(V_2-b{)}^{\gamma -1}=T_C(V_3-b{)}^{\gamma -1}⟹\frac{V_3-b}{V_2-b}={\left(\frac{T_H}{T_C}\right)}^{\frac{1}{\gamma -1}}$$$$T_H(V_1-b{)}^{\gamma -1}=T_C(V_4-b{)}^{\gamma -1}⟹\frac{V_4-b}{V_1-b}={\left(\frac{T_H}{T_C}\right)}^{\frac{1}{\gamma -1}}$$

Step 3: Equating the Volume Ratios

$$\frac{V_3-b}{V_2-b}=\frac{V_4-b}{V_1-b}⟹\frac{V_3-b}{V_4-b}=\frac{V_2-b}{V_1-b}$$

Step 4: Thermal Efficiency

$$\eta =1-\frac{Q_{\text{out}}}{Q_{\text{in}}}=1-\frac{R{T}_C\ln \left(\frac{V_3-b}{V_4-b}\right)}{R{T}_H\ln \left(\frac{V_2-b}{V_1-b}\right)}=\boxed{{\displaystyle 1-\frac{T_C}{T_H}={\eta }_{\text{Carnot}}}}$$

(Conclusion: The finite molecular volume $b$ shifts the operating volumes but leaves the fundamental Carnot efficiency invariant!)

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Volume 4: Modern Physics & Quantum Mechanics

Problem 4.1: Relativistic Compton Scattering with Moving Electrons

IPhO Olympiad Theoretical Masterclass

Problem Statement:
A high-energy photon of frequency ${\nu }_0$ collides head-on with an ultra-relativistic electron of rest mass $m_0$ moving in the opposite direction with velocity $\overrightarrow{v}=-\beta c\hat{i}$ ($\gamma =\frac{1}{\sqrt{1-{\beta }^2}}$). The photon is backscattered directly along its incident direction (${180}^{\circ }$ deflection). Derive the exact frequency ${\nu }^{\prime }$ of the backscattered photon.

<!-- Relativistic Electron --> <circle cx="340" cy="60" r="15" fill="#3b82f6" stroke="var(--vp-c-text-1)" stroke-width="1.5"/> <text x="334" y="65" fill="#fff" font-size="12" font-weight="700">e^-</text> <line x1="320" y1="60" x2="240" y2="60" stroke="#3b82f6" stroke-width="2.5" marker-end="url(#arrow-blue)"/> <text x="250" y="45" fill="#3b82f6" font-size="12" font-weight="700">Electron (&beta; c)</text> 

Figure Relativistic Inverse Compton Scattering: Head-on collision between incoming photon h*nu_0 and relativistic electron of speed beta*c.

💡 Click to view complete step-by-step solution

Step 1: Lorentz Transform into the Rest Frame of the Electron

In the electron's rest frame ($S^{\prime }$), the electron is stationary. The frame $S^{\prime }$ moves leftward with speed $v=\beta c$. By the Relativistic Doppler Shift, the incident photon has frequency ${\nu }_0^{\prime }$ in frame $S^{\prime }$:

$${\nu }_0^{\prime }={\nu }_0\sqrt{\frac{1+\beta }{1-\beta }}={\nu }_0\gamma (1+\beta )$$

Step 2: Standard Compton Scattering in Electron Rest Frame

In frame $S^{\prime }$, the electron is at rest, so standard Compton formula applies for scattering angle ${\theta }^{\prime }={180}^{\circ }$ ($\cos {180}^{\circ }=-1$):

$$\frac{1}{{\nu }^{‴}}-\frac{1}{{\nu }_0^{\prime }}=\frac{h}{m_0{c}^2}(1-\cos {180}^{\circ })=\frac{2h}{m_0{c}^2}$$$$\frac{1}{{\nu }^{‴}}=\frac{1}{{\nu }_0^{\prime }}+\frac{2h}{m_0{c}^2}=\frac{m_0{c}^2+2h{\nu }_0^{\prime }}{m_0{c}^2{\nu }_0^{\prime }}⟹{\nu }^{‴}=\frac{{\nu }_0^{\prime }}{1+\frac{2h{\nu }_0^{\prime }}{m_0{c}^2}}$$

Step 3: Transform Back to the Laboratory Frame

The backscattered photon moves in the $-x$ direction in $S^{\prime }$. Transforming back to the lab frame:

$${\nu }^{\prime }={\nu }^{‴}\sqrt{\frac{1+\beta }{1-\beta }}={\nu }^{‴}\gamma (1+\beta )$$

Substituting ${\nu }^{‴}$ and ${\nu }_0^{\prime }$:

$${\nu }^{\prime }=\frac{{\nu }_0\left(\frac{1+\beta }{1-\beta }\right)}{1+\frac{2h{\nu }_0\gamma (1+\beta )}{m_0{c}^2}}$$$$\boxed{{\displaystyle {\nu }^{\prime }={\nu }_0\left(\frac{1+\beta }{1-\beta }\right)\left[\frac{1}{1+\frac{2\gamma h{\nu }_0(1+\beta )}{m_0{c}^2}}\right]}}$$

(Physical Note: In Inverse Compton Scattering with $\gamma \gg 1$, the photon gains enormous energy, scaling as $\sim 4{\gamma }^2{\nu }_0$!)

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Problem 4.2: Positronium Ground State Annihilation Kinetics

JEE Advanced Multi-Concept

Problem Statement:
A positronium atom (a bound state of an electron $e^{-}$ and a positron $e^{+}$) has a ground state binding energy of $E_1=-6.8\text{ eV}$.

1. If the positronium atom annihilates at rest into two collinear gamma-ray photons, calculate the exact wavelength ${\lambda }_{\gamma }$ of each emitted photon.
2. If the atom had initial kinetic energy $K_0=10\text{ keV}$ before annihilation, find the maximum and minimum wavelengths of the emitted photons due to relativistic Doppler broadening.

💡 Click to view complete step-by-step solution

Step 1: Energy Conservation for Annihilation at Rest

The initial invariant mass energy is:

$$E_{\text{initial}}=2m_e{c}^2+E_1=2(511.0\text{ keV})-0.0068\text{ keV}=1021.9932\text{ keV}$$

By momentum conservation, the two photons travel in exactly opposite directions with equal energy:

$$E_{\gamma }=\frac{E_{\text{initial}}}{2}=m_e{c}^2+\frac{E_1}{2}=511.0\text{ keV}-3.4\text{ eV}\approx 510.9966\text{ keV}$$

Step 2: Wavelength of Emitted Photons

$${\lambda }_{\gamma }=\frac{hc}{E_{\gamma }}=\frac{12400\text{eV}\cdot \text{AA}}{510996.6\text{eV}}=\boxed{{\displaystyle 0.02426\text{AA}=2.426\text{ pm}}}$$

Step 3: Doppler Broadening with Kinetic Energy $K_0=10\text{ keV}$

Total energy of moving positronium: $E=M_0{c}^2+K_0$. The speed of the positronium is:

$$\gamma =1+\frac{K_0}{M_0{c}^2}=1+\frac{10\text{ keV}}{1022\text{ keV}}=1.00978⟹\beta =\sqrt{1-\frac{1}{{\gamma }^2}}\approx 0.1388$$

• Forward Emitted Photon (Blue-Shifted):

  $$E_{max}=E_{\gamma }\sqrt{\frac{1+\beta }{1-\beta }}=511\text{ keV}\times \sqrt{\frac{1.1388}{0.8612}}=511\times 1.1499=\boxed{{\displaystyle 587.6\text{ keV}}}$$$${\lambda }_{min}=\frac{hc}{E_{max}}=\boxed{{\displaystyle 0.0211\text{AA}}}$$

• Backward Emitted Photon (Red-Shifted):

  $$E_{min}=E_{\gamma }\sqrt{\frac{1-\beta }{1+\beta }}=511\times 0.8696=\boxed{{\displaystyle 444.4\text{ keV}}}$$$${\lambda }_{max}=\frac{hc}{E_{min}}=\boxed{{\displaystyle 0.0279\text{AA}}}$$