CBSE Class 12 Physics – Sample Paper Review Crack the pattern, conquer the paper.

Question Distribution & Your Strength

Chapter distribution vs. your proficiency map

Chapter # Questions Your Strength
Electric Charges & Fields 3 Low
Moving Charges & Magnetism 3 Low
Alternating Current 3 Low
Wave Optics 3 Low
Nuclei 2 Low
Electrostatic Potential & Capacitance 3 Medium
Magnetism & Matter 2 Medium
Electromagnetic Waves 2 Medium
Dual Nature of Radiation & Matter 2 Medium
Semiconductor Electronics 3 Medium
Current Electricity 4 High
Electromagnetic Induction 3 High
Ray Optics & Optical Instruments 3 High
Atoms 2 High

Goal: see how the paper maps to your syllabus strengths.

Question with Hints and Nudges

Class XII Physics – Electric Charges & Fields

Difficulty: Challenging
Self-Discovery

The Question:

A non-conducting spherical shell of radius \(R\) carries surface charge density \(\sigma(\theta)=\sigma_0\cos\theta\). Using Gauss’s law, find the electric field (magnitude & direction) (i) inside the shell \((r<R)\) and (ii) outside the shell \((r>R)\).

Spherical shell with \(\sigma=\sigma_0\cos\theta\)

Helpful Concepts

  • Gauss’s law: \( \Phi_E = Q_{\text{encl}}/\varepsilon_0 \).
  • Only enclosed charge influences flux through a closed surface.
  • \(\int \sigma_0\cos\theta\,dA=0\); the shell’s net charge is zero, resembling a dipole.
Gauss law Angular charge density

Progressive Hints (Reveal only when needed)

Gentle Nudge

Choose a spherical Gaussian surface of radius \(r\) concentric with the shell.

Direction Pointer

Evaluate \(Q_{\text{encl}}=\int \sigma_0\cos\theta\,dA\) using \(dA=R^2\sin\theta\,d\theta\,d\phi\).

Guiding Framework

The integral is zero for any \(r\). Hence flux is zero, so \(E=0\) inside. Outside, treat the shell as a dipole with moment \(p=\frac{4\pi R^{3}\sigma_0}{3}\) and use the standard dipole field.

Thinking Strategies

Break It Down

Find enclosed charge first; then relate \( \Phi_E \) to \(E\).

Connect to Prior Knowledge

Compare with a uniformly charged shell and with a point dipole.

Visualize It

Sketch positive and negative hemispheres to see the dipole pattern.

Test a Simpler Case

Replace \(\cos\theta\) with a constant to check your method gives Coulomb’s law.

Struggle is normal! Try to solve on your own before checking the hints.

Question with Hints and Nudges

CBSE Class 12 • Physics

Difficulty: Challenging
Self-Discovery

The Question:

A proton enters a uniform 0.20 T magnetic field with speed \(2\times10^{6}\,\text{m s}^{-1}\) at \(60^{\circ}\) to the field. Find (a) the radius of its circular path and (b) the pitch of the helix it traces.

Helpful Concepts

  • Lorentz force: \( \mathbf{F}=q(\mathbf{v}\times\mathbf{B}) \).
  • Only \(v_{\perp}\) causes circular motion; \(v_{\parallel}\) stays constant.
  • Pitch = distance moved along field in one revolution.
Lorentz force Pitch of helix

Progressive Hints (Reveal only when needed)

Gentle Nudge

Resolve the velocity into \(v_{\perp}\) and \(v_{\parallel}\) using sine and cosine of \(60^{\circ}\).

Direction Pointer

The Lorentz force bends only \(v_{\perp}\). Use \( r=\frac{m v_{\perp}}{qB} \) for the circle.

Guiding Framework

Calculate \( T=\frac{2\pi m}{qB} \). Then pitch = \( v_{\parallel} T \). Take \( m_p=1.67\times10^{-27}\,\text{kg} \), \( q_p=1.6\times10^{-19}\,\text{C} \).

Thinking Strategies

Break It Down

Treat perpendicular and parallel motions separately.

Connect to Prior Knowledge

Recall \(r=mv/qB\) from uniform circular motion in a magnetic field.

Visualize It

Draw the helical path around straight field lines.

Test a Simpler Case

Imagine \(60^{\circ}=90^{\circ}\) to check understanding of pure circular motion.

Struggle is normal! Try to solve on your own before checking the hints.

Question with Hints and Nudges

Class 12 Physics – Alternating Current

Difficulty: Hard
Self-Discovery

The Question:

An ideal series L-C-R circuit has L = 50 mH, C = 80 µF, R = 40 Ω and is driven by a 200 V, 50 Hz source. Find (a) the current amplitude and (b) the phase angle between current and source voltage.

Helpful Concepts

  • Inductive reactance: \(X_L = 2\pi f L\).
  • Capacitive reactance: \(X_C = \frac{1}{2\pi f C}\).
  • Impedance: \(Z = \sqrt{R^{2} + (X_L - X_C)^{2}}\); phase angle \(\tan\phi = \frac{X_L - X_C}{R}\).
Impedance Phase Angle

Progressive Hints (Reveal only when needed)

Gentle Nudge

Start by comparing \(X_L\) and \(X_C\) at 50 Hz. Which one is larger?

Direction Pointer

Numeric check: \(X_L \approx 15.7 Ω\), \(X_C \approx 39.8 Ω\). The circuit is capacitive.

Guiding Framework

Compute \(Z\) with the above values, use \(I_0 = V_0 / Z\). Find \(\tan\phi\); negative sign means current leads the voltage.

Thinking Strategies

Break It Down

Treat reactances, resistance and voltage separately before combining.

Connect to Prior Knowledge

Recall DC Ohm’s law—impedance plays the same role in AC.

Visualize It

Sketch the impedance triangle to see how \(R\) and \((X_L-X_C)\) combine.

Test a Simpler Case

Set C very large; notice how the circuit approaches a pure inductor–resistor pair.

Struggle is normal! Try to solve on your own before checking the hints.

Question with Hints and Nudges

Wave Optics – CBSE Class 12

Difficulty: Hard
Self-Discovery

The Question:

In a YDS set-up (λ = 600 nm, d = 0.5 mm, D = 1.5 m) a mica sheet (μ = 1.5) covers one slit so the central maximum shifts to the 3rd bright fringe. Find the sheet thickness t.

Helpful Concepts

  • YDS fringe width \( \beta = \frac{\lambda D}{d} \).
  • Sheet adds optical path \( \Delta = (\mu - 1)t \).
  • Central fringe shifts by \( m \) when \( \Delta = m\lambda \).
YDS Experiment Optical Path

Progressive Hints (Reveal only when needed)

Gentle Nudge

Which path difference makes the central bright coincide with the 3rd bright?

Direction Pointer

Compute \( \Delta = 3\lambda \) and link it to the sheet by \( (\mu - 1)t \).

Guiding Framework

Set \( (\mu - 1)t = 3\lambda \). With \( \lambda = 600\,\text{nm} \) and \( \mu = 1.5 \), you will get \( t = 3.6\,\mu\text{m} \).

Thinking Strategies

Break It Down

Isolate fringe shift first, then apply sheet optics.

Connect to Prior Knowledge

Recall how glass slabs create phase difference in interference.

Visualize It

Sketch fringe pattern before and after inserting mica.

Test a Simpler Case

Solve for \( m = 1 \) to check your method, then scale.

Struggle is normal! Try to solve on your own before checking the hints.

Question with Hints and Nudges

Nuclei – Fusion

Difficulty: Hard
Self-Discovery

The Question:

Two deuterons fuse: \(^{2}\text{H}+^{2}\text{H}\rightarrow{}^{3}\text{He}+n\). Given \(BE(^{2}\text{H}) = 1.113\ \text{MeV}\) and \(BE(^{3}\text{He}) = 7.718\ \text{MeV}\), compute the released energy \(Q\).

Helpful Concepts

  • Mass defect converts to binding energy: \( \Delta m c^{2}=BE \).
  • \(Q\)-value equals change in total binding energy.
  • 1 MeV ≈ \(1.602\times10^{-13}\) J for energy conversion.
Mass defect Energy conversion

Progressive Hints (Reveal only when needed)

Gentle Nudge

Add the binding energies of both reactant deuterons first.

Direction Pointer

The neutron is free, so its binding energy is zero; only \(^{3}\text{He}\) contributes for products.

Guiding Framework

Use \(Q = BE_{\text{products}} - BE_{\text{reactants}}\). A positive \(Q\) means energy is released.

Thinking Strategies

Break It Down

Compute reactant and product totals separately, then compare.

Connect to Prior Knowledge

Recall \(E=\Delta m c^{2}\): mass defect converts directly to energy.

Visualize It

Sketch energy levels; the deeper (more negative) level is more stable.

Test a Simpler Case

Try fusing one deuteron with a neutron to verify your calculation steps.

Question with Hints and Nudges

Electrostatic Potential & Capacitance

Difficulty: Challenging
Self-Discovery

The Question:

Three identical parallel-plate capacitors, each of capacitance \(C\), are connected in series to a 300 V source. While the circuit is live, the middle capacitor is completely filled with a dielectric of constant \(\kappa = 4\). Determine (i) the charge on every capacitor and (ii) the total energy stored in the combination.

Helpful Concepts

  • Series rule: \(1/C_{\text{eq}} = 1/C_1 + 1/C_2 + 1/C_3\).
  • Dielectric effect: \(C' = \kappa C\).
  • Energy stored: \(U = \tfrac12 C V^2\).
series capacitance energy storage

Progressive Hints (Reveal only when needed)

Gentle Nudge

Treat the middle capacitor as \(C' = 4C\) once the dielectric is inserted.

Direction Pointer

Find \(C_{\text{eq}}\) of the series set, then compute common charge \(Q = C_{\text{eq}} \times 300\text{ V}\).

Guiding Framework

Voltage drops: \(V_1 = Q/C\), \(V_2 = Q/C'\), \(V_3 = Q/C\). Check they sum to 300 V, then add energies: \(U = \tfrac12 \sum C_i V_i^2\).

Thinking Strategies

Break It Down

Analyze each capacitor, then combine their effects.

Connect to Prior Knowledge

Remember: charge is identical across capacitors in series.

Visualize It

Draw potential drops across the three plates.

Test a Simpler Case

Set \(\kappa = 1\) first; compare the result to see the dielectric’s impact.

Struggle is normal! Try to solve on your own before checking the hints.

Question with Hints and Nudges

CBSE Class 12 Physics

Difficulty: Hard
Self-Discovery

The Question:

A bar magnet oscillates freely in Earth’s horizontal field \(B_H = 3.6 \times 10^{-5}\,\text{T}\). Its time period is \(2\,\text{s}\).
(a) Find its magnetic moment \(M\).
(b) How far from the centre on the axial line will the net field be zero?

Diagram (if needed)

Helpful Concepts

  • Torsional oscillation of a magnetic dipole \(T = 2\pi\sqrt{\frac{I}{M B_H}}\)
  • Earth’s horizontal field acts like a uniform reference field
  • Axial field of a dipole \(B = \frac{\mu_0}{4\pi}\frac{2M}{r^3}\)
Oscillation method Null points

Progressive Hints (Reveal only when needed)

Gentle Nudge

Start with the oscillation relation linking \(T\), \(I\), \(M\) and \(B_H\).

Direction Pointer

Rearranged: \(M = \frac{4\pi^2 I}{B_H T^2}\). For a slender rod, \(I \approx m\ell^2/12\).

Guiding Framework

For part (b) set the dipole’s axial field opposite to Earth’s: \(\frac{\mu_0}{4\pi}\frac{2M}{r^3}=B_H\). Solve for \(r\).

Thinking Strategies

Break It Down

Tackle part (a) to get \(M\); use it immediately in part (b).

Connect to Prior Knowledge

Remember tangent law uses the same \(M/B_H\) ratio—consistent ideas help.

Visualize It

Sketch field lines; mark where the magnet’s axial field cancels Earth’s.

Test a Simpler Case

Assume \(M = 1\,\text{A·m}^2\) to verify your algebra before inserting numbers.

Electromagnetic Waves – Toughest Question

Physics – Class XII

Difficulty: Hard
Self-Discovery

The Question:

An EM wave with peak electric field \(E_0 = 6\;\text{V\,m}^{-1}\) travels along +x. (i) Write expressions for \(\mathbf{E}(x,t)\) and \(\mathbf{B}(x,t)\). (ii) Find (a) the magnetic field amplitude and (b) the average energy flux.

Helpful Concepts

  • Maxwell link: \(c = 1/\sqrt{\mu_0\varepsilon_0}\)
  • \(\mathbf{S} = \mathbf{E}\times\mathbf{B}/\mu_0\) → energy flux
  • Energy density \(u = \varepsilon_0 E^2\)
Maxwell Equations Energy Density

Progressive Hints (Reveal only when needed)

Gentle Nudge

Start with \(B_0 = E_0/c\) from Maxwell’s equations.

Direction Pointer

Choose \(\mathbf{E}\parallel\hat{j}\), \(\mathbf{B}\parallel\hat{k}\) so both are ⟂ to +x.

Guiding Framework

Write sine wave forms, compute \(B_0\), then use \(S_{\text{avg}} = E_0B_0/(2\mu_0)\).

Thinking Strategies

Break It Down

Treat amplitude, orientation, and energy flux as separate mini-tasks.

Connect to Prior Knowledge

Recall that \(c = 3\times10^{8}\,\text{m s}^{-1}\) links E and B fields.

Visualize It

Sketch mutually perpendicular \(\mathbf{E}\), \(\mathbf{B}\) and propagation axes.

Test a Simpler Case

Check with \(E_0=1\;\text{V m}^{-1}\) to avoid confusion between rms and peak values.

Struggle is normal! Try to solve on your own before checking the hints.

Question with Hints and Nudges

Dual Nature of Radiation & Matter

Difficulty: High
Self-Discovery

The Question:

Light of wavelength 248 nm falls on a metal with work function 2 eV.
(a) Calculate the maximum kinetic energy of the emitted electrons.
(b) What accelerating potential must be applied so their de Broglie wavelength becomes 0.50 Å?

Helpful Concepts

  • Photoelectric equation \(K_{\max}=E_{\gamma}-\phi\) (includes work function).
  • de Broglie wavelength \( \lambda=\frac{h}{\sqrt{2mK}} \).
  • Total kinetic energy after acceleration: \(K=K_{\max}+eV_{\text{acc}}\).
Work Function Kinetic Energy

Progressive Hints (Reveal only when needed)

Gentle Nudge

First find photon energy from its wavelength, then subtract the work function.

Direction Pointer

Convert \(E_{\gamma}=\frac{hc}{\lambda}\) to eV. Use \(K_{\max}=E_{\gamma}-2\text{ eV}\). For part (b), add \(eV_{\text{acc}}\) to \(K_{\max}\) before using the de Broglie formula.

Guiding Framework

Step 1: Compute \(K_{\max}\).
Step 2: Set \( \lambda =0.50\; \text{Å}=5.0\times10^{-11}\,\text{m}\).
Step 3: Solve \( \lambda =\frac{h}{\sqrt{2m(K_{\max}+eV_{\text{acc}})}} \) for \(V_{\text{acc}}\).

Thinking Strategies

Break It Down

Tackle part (a) completely before using its result for part (b).

Connect to Prior Knowledge

Recall how work function limits electron energy in photoelectric effect.

Visualize It

Sketch an energy bar showing photon energy, work function, \(K_{\max}\), and extra energy from acceleration.

Test a Simpler Case

Try λ = 1 Å first to see how potential affects wavelength before calculating the exact value.

Struggle is normal! Try to solve on your own before checking the hints.

Question with Hints and Nudges

Semiconductor Electronics – CE Amplifier

Difficulty: Hard
Self-Discovery

The Question:

For a silicon npn transistor in common-emitter mode, \(R_C = 2\,\text{k}\Omega\), \(R_E = 1\,\text{k}\Omega\) and \(\beta = 120\). A \(1\,\text{mV rms}\) input is fed through a \(1\,\text{k}\Omega\) base resistor.
(a) Find the small-signal voltage gain \(A_v\).
(b) Discuss the bias stability of this amplifier.

No additional diagram required

Helpful Concepts

  • Small-signal emitter resistance \(r_e \approx \frac{25\,\text{mV}}{I_C}\).
  • Voltage gain \(A_v \approx -\beta R_C / \bigl(r_e + (\beta+1)R_E\bigr)\).
  • Voltage-divider bias improves thermal stability of \(I_C\).
CE amplifier Bias stability

Progressive Hints (Reveal only when needed)

Gentle Nudge

Calculate \(I_b = \frac{1\,\text{mV}}{1\,\text{k}\Omega}\) and then find \(I_C = \beta I_b\).

Direction Pointer

Use \(r_e = 25\,\text{mV}/I_C\); expect \(r_e\) to be a few hundred ohms.

Guiding Framework

Insert \(r_e\) into \(A_v\). Note \((\beta + 1)R_E\) dominates the denominator, so \(A_v \approx -2\). Large \(R_E\) lowers gain but locks \(I_C\), giving excellent bias stability.

Thinking Strategies

Break It Down

Separate dc bias calculation from ac gain evaluation.

Connect to Prior Knowledge

Recall that emitter degeneration reduces gain but improves stability.

Visualize It

Sketch the small-signal model to see resistances in the signal path.

Test a Simpler Case

Ignore \(R_E\) briefly to observe how gain would rise to \(-\beta R_C/r_e\).

Question with Hints and Nudges

CBSE Class 12 Physics – Current Electricity

Difficulty: Hard
Self-Discovery

The Question:

A uniform wire of resistance \(2\,\Omega\) is bent into a square. Determine the resistance between opposite corners. The wire is then stretched so one side becomes twice its original length, volume constant. Find the new resistance between the same opposite corners.

Sketch the square and label resistances.

Helpful Concepts

  • \(R = \rho \dfrac{l}{A}\); stretch keeps volume \(lA\) constant.
  • Balanced Wheatstone bridge ⇒ no current through diagonal.
  • Series – parallel reduction simplifies symmetric networks.
Wheatstone bridge Meter bridge

Progressive Hints (Reveal only when needed)

Gentle Nudge

First, compute resistance between corners when each side is \(0.5\,\Omega\).

Direction Pointer

Treat the square as a balanced Wheatstone bridge; ignore the diagonal branch.

Guiding Framework

Stretching doubles one side’s length. Remaining three sides shorten equally. Use \(R \propto l^2\) then re-apply the bridge reduction.

Thinking Strategies

Break It Down

Analyse pre-stretch and post-stretch networks separately.

Connect to Prior Knowledge

Recall meter-bridge balance ⇒ equal potential nodes cancel internal branch.

Visualize It

Sketch potentials on the square; mark equipotential corners.

Test a Simpler Case

Imagine stretching until three sides vanish; check if trend matches answer.

Struggle is normal! Try to solve on your own before checking the hints.

Question with Hints and Nudges

Class XII Physics – Electromagnetic Induction

Difficulty: Hard
Self-Discovery

The Question:

A circular coil of 200 turns and area 0.02 m² rotates at 300 rpm about its diameter in a uniform 0.04 T field. (a) Derive the expression for the instantaneous induced emf. (b) Calculate the peak emf produced.

Visualise the rotating coil of an AC generator.

Helpful Concepts

  • Faraday’s law: \( \varepsilon = -N \frac{d\Phi}{dt} \).
  • Flux for rotating loop: \( \Phi = BA\cos\omega t \).
  • Peak emf of an AC generator: \( \varepsilon_{0}=N B A \omega \).
Induced emf AC generator

Progressive Hints (Reveal only when needed)

Gentle Nudge

Start by writing the magnetic flux through one turn as a time-dependent cosine.

Direction Pointer

Differentiate \( \Phi \) with respect to \( t \) to get \( \varepsilon \); then multiply by 200 turns.

Guiding Framework

Convert 300 rpm to \( \omega = 2\pi n \). Use \( \varepsilon_{0}=N B A \omega \) to find \( \approx 5.0\; \text{V} \).

Thinking Strategies

Break It Down

First derive the formula, then plug numbers.

Connect to Prior Knowledge

Recall how generators convert rotation to alternating voltage.

Visualize It

Sketch coil positions for \( \cos \) and \( \sin \) angles.

Test a Simpler Case

Analyze a single-turn loop, then scale by 200.

Struggle is normal! Try to solve on your own before checking the hints.

Question with Hints and Nudges

Ray Optics & Optical Instruments

Difficulty: Hard
Self-Discovery

The Question:

A refracting telescope has an objective of focal length 80 cm and an eyepiece of focal length 4 cm. The final image is at infinity.
(a) Find the lens separation.
(b) Determine the angular magnification.
(c) What objective diameter gives 200× the light-gathering power of a 6 mm pupil?

Helpful Concepts

  • Normal adjustment: \(d \approx f_0 + f_e\).
  • Angular magnification \(M = \frac{f_0}{f_e}\).
  • Light-gathering power \(\propto D^2\).
Telescope Angular Magnification

Progressive Hints (Reveal only when needed)

Gentle Nudge

In normal adjustment, the objective image lies at the eyepiece focal plane.

Direction Pointer

Use \(d = f_0 + f_e\) for separation and \(M = f_0 / f_e\) for magnification.

Guiding Framework

Calculate \(M = 80/4\). For light power, set \(\left(\frac{D_0}{6\text{ mm}}\right)^2 = 200\) and solve for \(D_0\).

Thinking Strategies

Break It Down

Solve parts (a), (b), and (c) one after another.

Connect to Prior Knowledge

Link magnification to focal-length ratios you learned for simple lenses.

Visualize It

Draw a ray diagram of the telescope in normal adjustment.

Test a Simpler Case

Try equal focal lengths to check if your formulas give \(M = 1\).

Atoms – Toughest Question

Atoms • Hydrogen Spectrum

Difficulty: Hard
Self-Discovery

The Question:

Using Bohr postulates, derive the wavelength expression for the Balmer series (transitions to \(n_1 = 2\)) and calculate its shortest-wavelength limit.

Helpful Concepts

  • Bohr energy levels: \(E_n = -13.6\,\text{eV}/n^2\).
  • Photon energy–wavelength link: \( \Delta E = hc/\lambda \).
  • Rydberg formula: \(1/\lambda = R\,(1/n_1^2 - 1/n_2^2)\).
Energy Levels Rydberg Formula

Progressive Hints (Reveal only when needed)

Gentle Nudge

Write \( \Delta E = 13.6\,\text{eV}\,(1/n_1^2 - 1/n_2^2) \) for the photon released.

Direction Pointer

Substitute \(n_1 = 2\). Use \( \Delta E = hc/\lambda \) to obtain \(1/\lambda\).

Guiding Framework

Let \(n_2 \to \infty\). Then \(1/\lambda_{\min} = R/4\), so \( \lambda_{\min} \approx 364.6\,\text{nm} \).

Thinking Strategies

Break It Down

Identify initial and final levels, compute \( \Delta E \), then convert to \( \lambda \).

Connect to Prior Knowledge

Remember Balmer lines lie in the visible region; check if your answer fits.

Visualize It

Sketch energy levels and draw arrows for transitions to \( n=2 \).

Test a Simpler Case

Calculate \( \lambda \) for \( n_2 = 3 \) first to verify your method.

Struggle is normal! Try to solve on your own before checking the hints.

Key Take-aways

Check Symmetry First

Quick recap—verify symmetry, then apply \( \oint \vec{E}\!\cdot\!d\vec{A}=q_{\text{enc}}/\varepsilon_0 \).

Split Velocity Components

Resolve \( \vec{v} \) into parts ⟂ and ∥ \( \vec{B} \) for instant path insight.

Compare \(X_L\) & \(X_C\)

Judge lead/lag and impedance size before any AC number-crunching.

Measure Path Difference

Use \( \Delta = d\sin\theta \) to locate fringes swiftly in interference.

Find Δm First

Compute mass or binding-energy change, then apply \(E=\Delta m c^{2}\).

Include \( r_e \) in Model

Add emitter resistance in BJT small-signal diagrams for accurate gain.

Mind Your Units

Last-minute things to remember: write rad, J, T clearly to avoid slips.