Comprehensive Physics Maha Marathon Review for IAT/NEST 2026

Physics Marathon Review: Core Strategies and Exam Patterns

• Target Exams: This comprehensive review is tailored for NEST and IAT 2026, targeting admissions to IISc, IISERs, and NISER. It integrates full Class 12th revision with high-level problem-solving relevant to JEE-Advanced style questions.

• Exam Strategy: Direct Question Mapping: Evidence suggests that NEST and IISER Aptitude Test (IAT) frequently repeat questions from JEE-Advanced papers dating back 10-20 years.

  • Case 1: JEE-Advanced 2012 (Cylinder with cavity) appeared verbatim in IAT 2022.

  • Case 2: JEE-Advanced 1999 (Rolling disk angular momentum) was copied into NEST 2018.

  • Case 3: JEE-Advanced 2014 (Ladder equilibrium) appeared in IIT JEE 2020 with a flipped diagram.

• General Strategy: Solving JEE-Advanced problems is critical not only for concept building but because they serve as a direct primary source for other elite competitive exams.

Electrostatics: Force, Fields, and Superposition Analysis

• Proton Motion Between Parallel Plates (JEE-Advanced 2012):

  • Scenario: Two large vertical plates with separation $d = 0.01\,m$ connected to a voltage source. A proton ($m_p = 1.67 \times 10^{-27}\,kg$, $q = 1.6 \times 10^{-19}\,C$) is released from rest at the midpoint and moves at an angle of $45^{\circ}$ to the vertical.

  • Motion Principle: Direction of motion initially follows the direction of the net force when starting from rest. Net force ($\mathbf{F}_{net}$) has components: vertical ($mg$) and horizontal ($qE$).

  • Balance Analysis: Since the angle is $45^{\circ}$, the magnitudes of the forces must be equal: $\tan(45^{\circ}) = \frac{qE}{mg} = 1$. Therefore, $qE = mg$.

  • Electric Field and Voltage: $E = \frac{mg}{q}$. In a parallel plate setup, $V = E \times d$.

  • Numerical Result: $V = \frac{1.67 \times 10^{-27} \times 10 \times 0.01}{1.6 \times 10^{-19}} \approx 10^{-9}\,V$. The order of magnitude of current/potential here relates to specific orders of $V$.

• The Superposition Principle for Cavities:

  • Principle: The field from an object with a cavity equals the field from the solid object minus the field from the cavity region filled with same charge density.

  • Cylinder with Spherical Cavity:

    • Inner cylinder radius: $R$. Cavity center on axis.

    • Solid cylinder field at $r = 2R$: $E_{cylinder} = \frac{\rho R^2}{4\epsilon_0}$.

    • Spherical cavity contribution (treating as sphere of negative charge): $E_{sphere} = \frac{1}{4\pi\epsilon_0} \frac{Q_{cavity}}{(2R)^2}$, where $Q_{cavity} = -\rho \times \frac{4}{3}\pi (R/2)^3$.

    • Final result yields a magnitude of field at P approximately $\frac{23\rho R}{96\epsilon_0}$.

  • Uniform Sphere Cavity (JEE-Advanced 2015):

    • Result: The electric field inside the cavity of a uniformly charged sphere is always uniform (constant magnitude and direction).

    • Vector identity: $\mathbf{E}_{cavity} = \frac{\rho\mathbf{a}}{3\epsilon_0}$, where $\mathbf{a}$ is the vector from the center of the sphere to the center of the cavity.

Flux, Solid Angles, and Symmetry in Electrostatics

• The Hemispherical Bowl (JEE-Advanced 2017):

  • Setup: Point charge $+q$ placed infinitesimally above the center of a hemispherical bowl of radius $r$.

  • Gauss's Law: Since the charge is just outside the closed surface (bowl + flat lid), the net flux $\Phi_{total} = 0$.

  • Flux Distribution: $\Phi_{net} = \Phi_{curved} + \Phi_{flat} = 0 \implies \Phi_{curved} = -\Phi_{flat}$.

  • Solid Angle Method:

    • Formula for flux: $\Phi = \frac{q}{4\pi\epsilon_0} \Omega$.

    • For a cone with half-angle $\theta$: $\Omega = 2\pi(1 - \cos(\theta))$.

    • Geometry here: In the cone from charge to the rim, $\theta = 45^{\circ}$ because the vertical distance $r$ and horizontal radius $r$ are equal. $\cos(45^{\circ}) = \frac{1}{\sqrt{2}}$.

    • Therefore, $\Phi_{flat} = \frac{q}{2\epsilon_0} (1 - \frac{1}{\sqrt{2}})$.

  • Equipotential Geometry: All points on the circumference (rim) are equidistant ($\sqrt{r^2 + r^2} = r\sqrt{2}$) from the charge; thus, the circumference is an equipotential line.

• Regular Hexagon Configuration (JEE-Advanced 2012):

  • Six charges at vertices of side $L$. Center $O$, defined constant $K = \frac{q}{4\pi\epsilon_0 L^2}$.

  • Electric Field: Vector sum. If charges are asymmetrical (e.g., $+2q, -q, +q, -2q, +q, -q$), the net field must be calculated by pairing vectors. At the center, current results show $E_{net} = 6K$ along the downward vertical axis.

  • Potential: Scalar sum. If $\sum q_i = 0$, potential at geometric center is $0$. On the axis of symmetry $PR$, if charges are mirrored with opposite signs, potential is zero everywhere on the line.

Dynamics of Charges and Capacitors

• Conducting Balls in a Parallel Plate Capacitor (JEE-Advanced 2015):

  • System: Two plates at potential $+V_0$ and $-V_0$. Lightweight balls with conducting coating (likened to silver-coated laddus) placed inside.

  • Mechanism:

    1. Ball touches bottom plate ($+V_0$) and becomes positively charged.

    2. $\mathbf{F} = q\mathbf{E}$ repels it toward the top plate.

    3. Collision is perfectly inelastic ($e = 0$); ball sticks to top plate momentarily, exchanges charge to $-q$, and is then repelled downward.

  • Motion: The motion is periodic but NOT Simple Harmonic Motion (SHM) because the restoring force is not proportional to displacement ($F = constant$, not $F = -kx$).

  • Current Relationship: Average current $I = \frac{q}{T}$. With $q = CV_0$ and travel time $t \propto \sqrt{\frac{m}{qE}}$, the steady-state current $I$ is proportional to $V_0^{3/2}$.

• Capacitor with Multi-Dielectric Layers:

  • Arrangements with half-fill dielectrics create series and parallel combinations.

  • For layers in series: $\frac{1}{C_{series}} = \frac{1}{C_3} + \frac{1}{C_4}$.

  • Total capacitance relates to area ratio and relative permittivity $\epsilon_r$.

Magnetostatics and Magnetic Field Properties

• Force on Arbitrary Conductors:

  • In a uniform magnetic field $\mathbf{B}$, the total force $\mathbf{F} = I\mathbf{L}_{eff} \times \mathbf{B}$, where $\mathbf{L}_{eff}$ is the vector displacement from the start point to the end point.

  • Example: A semicircular wire of radius $R$ and a straight wire of length $2R$ experience the same force in a uniform field if their endpoints coincide.

• Hollow Cylindrical Conductor Field Analysis:

  • Inner radius $r/2$, outer radius $r$, current density $J$.

  • Region 1 (R < r/2): Inside the central hollow, $I_{encl} = 0 \implies B = 0$.

  • Region 2 (r/2 < R < r): $B(2\pi R) = \mu_0 J(\pi[R^2 - (r/2)^2])$. The field increases non-linearly.

  • Region 3 (R > r): Outside, the whole cylinder acts like a thin wire carrying the total current. $B \propto \frac{1}{R}$.

• Magnetic Moment of Composite Loops:

  • Loop consists of 4 semicircular arcs of radius $a$ and a central square of side $a$.

  • Calculation: Equivalent area is two full circles ($2\pi a^2$) plus the square ($a^2$).

  • $\mathbf{m} = I \times A_{total} = I a^2(2\pi + 1)\mathbf{k}$.

Electromagnetic Induction (EMI) and AC Resonance

• Rotating Loops and Flux Cancellation:

  • Two loops with areas $A$ and $2A$ share a common axis but are on opposite sides.

  • When rotating in field $\\mathbf{B}$, their area vectors are antiparallel. Net flux $\Phi_{net} = \Phi_1 + \Phi_2 = BA\cos(\omega t) - 2BA\cos(\omega t) = -BA\cos(\omega t)$.

  • Induced EMF: $\mathcal{E} = -d\Phi/dt = -BA\omega \sin(\omega t)$. The maximum EMF is determined by the difference in areas.

• LCR Circuit Characteristics:

  • Resonance: Occurs when $X_L = X_C$. Resonant frequency $\omega_0 = \frac{1}{\sqrt{LC}}$.

  • Independence: $\omega_0$ depends solely on $L$ and $C$, not on resistance $R$. $R$ only affects the Q-factor (sharpness).

  • Frequency Limits:

    • As $\omega \rightarrow 0$, $X_C \rightarrow \infty$, current $I \rightarrow 0$ (capacitor blocks DC).

    • As $\omega \rightarrow \infty$, $X_L \rightarrow \infty$, current $I \rightarrow 0$ (inductor blocks high frequencies).

  • For $L = 1\,mH$, $C = 1\,\mu F$, $\omega_0 = \frac{1}{\sqrt{10^{-3} \times 10^{-6}}} = 10^{4.5} \approx 31622\,rad/s$.

Modern Physics: Quanta and Atoms

• De Broglie Wavelength ($\lambda_d$):

  • Formula: $\lambda_d = \frac{h}{p} = \frac{h}{\sqrt{2mKE}}$.

  • Proton scattering experiment: At closest approach $r$, $KE = PE_{electric}$. $\lambda_d = \frac{h}{\sqrt{2m [\frac{k q_1 q_2}{r}]}}$. For a proton approaching a nucleus at $10\,fm$, $\lambda_d \approx 7\,fm$.

• Einstein's Photoelectric Equation and Graphs:

  • $eV_0 = \frac{hc}{\lambda} - \phi_0$.

  • Graph of stopping potential $V_0$ vs. $1/\lambda$ is linear with slope $\frac{hc}{e}$.

  • Graph of $\lambda_d$ (photoelectron) vs. wavelength $\lambda$: $\lambda_d = \frac{h}{\sqrt{2m(\frac{hc}{\lambda} - \phi_0)}}$. Taking the ratio $\Delta \lambda_d / \Delta \lambda$ shows dependence on $\lambda^2 \lambda_d^{-3}$.

• Atomic Transitions (Hydrogen Atom):

  • Energy: $E_n = -\frac{13.6}{n^2}\,eV$. Potential Energy $V_n = 2E_n = -\frac{27.2}{n^2}\,eV$.

  • Ratio derived from $V_f / V_i = 6.25$ implies $(n_i/n_f)^2 = 6.25 \implies n_i/n_f = 2.5$. Smallest integers: $n_i = 2, n_f = 5$.

  • Spectral Lines: Transition from $n=4$ to ground state yields $N = \frac{n(n-1)}{2} = \frac{4 \times 3}{2} = 6$ spectral lines.

• Radioactivity and Mean Life:

  • Mean life $\tau = 1/\lambda$. Activity $A = \lambda N = \frac{N}{\tau}$. Rate of change $R = \frac{dA}{dt} = -\lambda A = -\frac{A}{\tau}$.

  • Ratio of rates of change $R_Q/R_P$ at $t = 2\tau$ is calculated using initial equal activities. Final value involves factors of $e$.

Geometrical and Wave Optics

• Compound Optical Systems:

  • Lens in Medium: Focal length changes according to Lens Maker's Formula: $\frac{1}{f'} = (\frac{\mu_l}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.

  • Mirror-Lens Combination with Tilted Mirror: Requires rotating coordinate systems. If a mirror is tilted at $30^{\circ}$, its local x-axis is rotated. Image position calculated in local frame must be transformed back to lab frame using a rotation matrix.

• Variable Refractive Index ($n(y)$):

  • For a slab where $n$ increases with depth $y$, light follows a curved path.

  • Extended Snell's Law: $n_1\sin(\theta_i) = n(y)\sin(\theta(y)) = n_2\sin(\theta_{emerge})$.

  • Lateral displacement $l$ depends on the gradient $dn/dy$ and slab thickness, but not directly on the final medium's index if the boundary is parallel.

• 3D Young's Double Slit Experiment (YDSE):

  • When the screen is placed perpendicular to the plane of the slits (XZ-plane), the symmetry about the line joining the sources results in semi-circular fringes centered on the origin $O$, rather than straight lines or hyperbolas in standard 2D setups.