Preparing for the CBSE Class 12 Physics board exam can feel overwhelming given the vast syllabus, but focusing on the right topics and practicing the right question formats makes all the difference. This article zeroes in on the most important topics from each unit—concepts like optics, electrostatics, magnetism, and modern physics—specifically for the Class 12 board exam. With MCQs now accounting for 30–40% of the 70-mark theory paper, the student who masters chapter-wise multiple choice questions has a massive advantage over one who only practises long-form derivations.
Yet most students make the same costly mistake: they solve random questions from random sources, get no feedback on why they were wrong, and repeat the same errors right up to exam day. This guide fixes that. Below you will find every chapter of the CBSE Class 12 Physics syllabus broken down into high-yield MCQ topics, 2–3 carefully chosen questions with full explanations, and — most importantly — the exact common mistake that trips students up on each one. Work through this chapter by chapter, and you will not just memorise answers; you will understand Physics well enough to handle any MCQ the board throws at you.
Why chapter-wise MCQs are critical for Physics
MCQs now carry 30–40% weightage in the CBSE Class 12 Physics board paper. With 9 units and 14 chapters in the syllabus, marks are distributed unevenly — Electrostatics, Optics, and Dual Nature appear almost every year. Chapter-wise practice helps you target weak areas, decode question patterns, and walk into the exam confident.
👉 Most students repeatedly solve the same easy numericals and miss concept-based MCQs entirely
👉 No clear explanation for why a particular option is wrong — leading to repeated mistakes
👉 No systematic way to track which chapters are truly exam-ready
Unit I · 16 Marks
Electrostatics
Chapters 1–2
01
Electric Charges and Fields
High yieldNEET
High-yield MCQ topics
Coulomb's law & superposition
Electric field due to dipole
Gauss's theorem applications
Electric field lines properties
Torque on a dipole
MCQ 1Which of the following statements about electric field lines is incorrect?
(a) They start from positive charges
(b) They end on negative charges
(c) Two field lines can cross each other ✓
(d) They are perpendicular to equipotential surfaces
Answer: (c)
Electric field lines can never intersect. If two lines crossed, it would imply two directions of the electric field at a single point — which is physically impossible since the field has a unique direction at every point.
Students often miss this conceptual trap. Always verify each property of field lines — the crossing condition is the most frequently tested incorrect statement.
MCQ 2The electric field at a point on the equatorial line of an electric dipole is:
(a) Along the dipole moment direction
(b) Opposite to the dipole moment direction ✓
(c) Perpendicular to the dipole axis
(d) Zero
Answer: (b)
On the equatorial line, the field is antiparallel to the dipole moment p̂. On the axial line, it is parallel to p̂. This direction reversal is a classic 1-mark board question.
Axial field → parallel to p̂. Equatorial field → antiparallel to p̂. Confusing these two directions is one of the most common errors in this chapter.
MCQ 3According to Gauss's law, the net electric flux through any closed surface equals:
(a) Total charge outside the surface / ε₀
(b) Total charge enclosed by the surface / ε₀ ✓
(c) Surface area × electric field
(d) Charge on the surface / ε₀
Answer: (b)
Φ = Q_enc / ε₀. Only the enclosed charge contributes to the net flux. External charges affect field distribution but contribute nothing to the net flux through the closed surface.
Always specify "enclosed" — external charges have zero net contribution to flux even though they affect the local field.
02
Electrostatic Potential and Capacitance
High yieldBoard priority
High-yield MCQ topics
Equipotential surfaces
Capacitors in series & parallel
Energy stored in a capacitor
Effect of dielectric slab
Potential due to a dipole
MCQ 1When a dielectric slab is inserted between the plates of an isolated (disconnected) charged capacitor:
(a) Charge increases
(b) Voltage increases
(c) Capacitance increases; voltage decreases ✓
(d) Energy stored increases
Answer: (c)
For an isolated capacitor, charge Q is fixed. Inserting a dielectric increases C by factor K. Since V = Q/C, voltage decreases. Energy U = Q²/2C also decreases — the energy goes into polarising the dielectric material.
Critical distinction: isolated capacitor (Q fixed) vs connected-to-battery capacitor (V fixed). Results are opposite for each case — this comparison is a very common multi-part MCQ trap.
MCQ 2The work done in moving a test charge along an equipotential surface is:
(a) Zero ✓
(b) Maximum
(c) Depends on the path taken
(d) Equal to the potential at that surface
Answer: (a)
Potential difference between any two points on an equipotential surface is zero, so W = qΔV = 0. Also, E is always perpendicular to equipotential surfaces, meaning no force component acts along the surface.
Link these three: equipotential surface → ΔV = 0 → W = 0 → E ⊥ surface. All three are tested together in assertion-reason questions.
Unit II · Part of 16 Marks
Current Electricity
Chapter 3
03
Current Electricity
High yieldNEET
High-yield MCQ topics
Drift velocity & mobility
Ohm's law & V-I characteristics
Kirchhoff's laws
Wheatstone bridge balance
Internal resistance & EMF
Temperature dependence of resistance
MCQ 1In a Wheatstone bridge, the bridge is said to be balanced when:
(a) All four resistances are equal
(b) P/Q = R/S and no current flows through galvanometer ✓
(c) Current through the battery is zero
(d) Voltage across R and S are equal
Answer: (b)
The balance condition is P/Q = R/S, meaning no current flows through the galvanometer. Equal resistances satisfy this but the general condition does not require all four to be equal.
Don't fall for option (a). Equal resistances are a special case, not the general balance condition. Know the ratio form: P/Q = R/S.
MCQ 2For n identical cells connected in parallel, the effective EMF and internal resistance are:
(a) nε and nr
(b) ε and r/n ✓
(c) ε/n and r/n
(d) nε and r/n
Answer: (b)
For identical cells in parallel, EMF remains equal to a single cell (ε), but internal resistance reduces to r/n. Parallel combination increases current capacity, not voltage. For series: EMF = nε, internal resistance = nr.
Series → EMF adds (nε), internal resistance adds (nr). Parallel → EMF stays same (ε), internal resistance reduces (r/n). Confusing these is one of the most common calculation errors.
Unit III · 17 Marks
Magnetic Effects of Current and Magnetism
Chapters 4–5
04
Moving Charges and Magnetism
High yieldNEET
High-yield MCQ topics
Biot-Savart law applications
Ampere's law for solenoid
Force on moving charge (r = mv/qB)
Galvanometer → ammeter/voltmeter
Torque on current loop
MCQ 1A proton and an alpha particle enter a uniform magnetic field with the same velocity. The ratio of radii of their circular paths (r_p : r_α) is:
(a) 1 : 2 ✓
(b) 2 : 1
(c) 1 : 4
(d) 1 : 1
Answer: (a) 1:2
r = mv/qB, so r ∝ m/q for same v and B. Proton: m_p/e. Alpha: 4m_p/2e = 2m_p/e. Therefore r_p : r_α = (m_p/e) : (2m_p/e) = 1:2.
Alpha particle has mass 4m_p but charge 2e — students often mistakenly use charge e for alpha, leading to an incorrect ratio of 1:4.
MCQ 2To convert a galvanometer into an ammeter, one must connect:
(a) A low resistance (shunt) in parallel ✓
(b) A high resistance in series
(c) A low resistance in series
(d) A high resistance in parallel
Answer: (a)
A shunt (small resistance in parallel) diverts most of the current away from the galvanometer coil, enabling measurement of large currents. A high resistance in series converts it to a voltmeter instead.
Ammeter = shunt (low R, parallel). Voltmeter = high R in series. This distinction is tested almost every year in 1-mark questions.
05
Magnetism and Matter
Board priority
High-yield MCQ topics
Para, dia, ferromagnetic substances
Bar magnet as solenoid
Curie's law & temperature effect
Magnetic susceptibility
MCQ 1Which of the following materials is ferromagnetic?
(a) Aluminium
(b) Bismuth
(c) Iron ✓
(d) Copper
Answer: (c)
Iron (Fe) is ferromagnetic — it has very high susceptibility and is strongly attracted to magnets. Aluminium is paramagnetic (weakly attracted); Bismuth and Copper are diamagnetic (weakly repelled).
Diamagnetic: Bi, Cu, water, gold. Paramagnetic: Al, O₂, Pt, Na. Ferromagnetic: Fe, Ni, Co. Memorise at least two examples from each class.
Unit IV · Part of 17 Marks
Electromagnetic Induction and Alternating Currents
Chapters 6–7
06
Electromagnetic Induction
High yieldNEET
High-yield MCQ topics
Faraday's laws of induction
Lenz's law direction
Self & mutual inductance
Induced EMF in a moving rod
MCQ 1Lenz's law is a consequence of the law of conservation of:
(a) Charge
(b) Mass
(c) Momentum
(d) Energy ✓
Answer: (d)
Lenz's law states that the induced current opposes the change in flux that causes it. This opposition means external work must be done against the induced force — a direct consequence of energy conservation.
The key word in Lenz's law is "opposes." Opposition requires doing work, which is why energy conservation is the underlying principle.
MCQ 2The SI unit of self-inductance is:
(a) Weber
(b) Tesla
(c) Henry ✓
(d) Farad
Answer: (c)
Self-inductance L is measured in Henry (H). 1 H = 1 V·s/A. Inductance is the electrical analogue of inertia — a coil with higher L resists changes in current more strongly.
Units confusion is common: Capacitance → Farad; Inductance → Henry; Magnetic flux → Weber; Magnetic field → Tesla. Don't mix them up in 1-mark questions.
07
Alternating Current
High yieldBoard priority
High-yield MCQ topics
RMS vs peak values
LCR series resonance
Power factor & wattless current
AC generator & transformer
Reactance of L and C
MCQ 1At resonance in a series LCR circuit, the impedance is:
(a) Maximum and equal to ωL
(b) Minimum and equal to R ✓
(c) Zero
(d) Equal to (X_L − X_C)
Answer: (b)
At resonance X_L = X_C, so they cancel. Impedance Z = √(R² + (X_L−X_C)²) reduces to Z = R, the minimum value. Current is therefore maximum at resonance.
At resonance: Z = R (min), I = max, power factor cos φ = 1, V_L = V_C. All four properties are individually tested in boards — know all of them.
Unit V · 18 Marks
Electromagnetic Waves
Chapter 8
08
Electromagnetic Waves
Board priority
High-yield MCQ topics
EM spectrum order & wavelengths
Uses of each EM wave type
Displacement current (Maxwell)
Transverse nature of EM waves
MCQ 1Which electromagnetic radiation is used in RADAR systems?
(a) Infrared
(b) X-rays
(c) Microwaves ✓
(d) Ultraviolet
Answer: (c)
Microwaves (~GHz frequency) are used in RADAR (Radio Detection And Ranging) and microwave ovens. IR → remote controls; UV → sterilisation; X-rays → medical imaging; Gamma → cancer treatment.
Learn at least one application for each EM wave type. This is a pure fact-based MCQ — no formula needed, just accurate recall.
MCQ 2Displacement current exists when:
(a) Steady current flows in a conductor
(b) A static charge is present
(c) Electric flux through a surface is changing with time ✓
(d) Magnetic field is constant
Answer: (c)
Maxwell introduced displacement current I_d = ε₀ dΦ_E/dt, which arises from a changing electric flux — for example, inside a charging capacitor where no actual charge flows between the plates.
Displacement current is not a real flow of charges — it is a term added by Maxwell to complete Ampere's law and make it consistent with the continuity equation.
Unit VI · Part of 18 Marks
Optics
Chapters 9–10 · Highest MCQ density section
09
Ray Optics and Optical Instruments
High yieldNEET
High-yield MCQ topics
Mirror & lens formula
Total internal reflection
Lens maker's formula
Power of lens combination
Refraction through a prism
Optical fibre principle
MCQ 1The critical angle for total internal reflection depends on:
(a) Only the angle of incidence
(b) The refractive indices of the two media and wavelength of light ✓
(c) Intensity of incident light only
(d) Only the denser medium
Answer: (b)
sin C = n₂/n₁ (for light going from medium 1 to medium 2, n₁ > n₂). Since n depends on wavelength, different colours have different critical angles. Violet light has a smaller critical angle than red light.
TIR requires: (1) light travels from denser to rarer medium, AND (2) angle of incidence ≥ critical angle. Both conditions must be satisfied simultaneously.
MCQ 2Two thin lenses of focal lengths f₁ and f₂ are placed in contact. The power of the combination is:
(a) f₁ + f₂
(b) f₁ × f₂
(c) 1/f₁ + 1/f₂ ✓
(d) 1/(f₁ + f₂)
Answer: (c)
P = P₁ + P₂ = 1/f₁ + 1/f₂. Equivalently, 1/f = 1/f₁ + 1/f₂ for thin lenses in contact. Powers add directly; focal lengths add as reciprocals.
Powers add: P = P₁ + P₂. Focal lengths combine as 1/f = 1/f₁ + 1/f₂. Never add focal lengths directly — that is the most common error in this topic.
10
Wave Optics
High yieldNEET
High-yield MCQ topics
YDSE fringe width β = λD/d
Constructive & destructive interference
Huygen's principle
Single slit diffraction pattern
Coherent sources
MCQ 1In Young's double slit experiment, if the slit separation is doubled and the distance to the screen is halved, fringe width becomes:
(a) 4 times the original
(b) 1/4 of the original ✓
(c) Unchanged
(d) Half of the original
Answer: (b)
β = λD/d. New β' = λ(D/2)/(2d) = λD/4d = β/4. When d doubles and D halves, the fringe width becomes one-fourth of the original value.
Always substitute into β = λD/d rather than trying to remember individual cases. Fringe width ∝ D and ∝ λ; inversely ∝ d.
MCQ 2The phenomenon of interference is based on the principle of:
(a) Reflection
(b) Refraction
(c) Superposition of waves ✓
(d) Diffraction
Answer: (c)
Interference arises when two coherent waves superpose. Path difference = nλ → constructive (bright fringe); path difference = (2n−1)λ/2 → destructive (dark fringe).
Interference requires coherent sources (constant phase difference). Two independent light sources cannot produce a stable interference pattern.
Unit VII · 12 Marks
Dual Nature of Radiation and Matter
Chapter 11
11
Dual Nature of Radiation and Matter
High yieldNEET
High-yield MCQ topics
Photoelectric effect conditions
Einstein's photoelectric equation
Work function & threshold frequency
de-Broglie wavelength (λ = h/p)
Stopping potential
MCQ 1When the intensity of incident light is doubled (frequency kept above threshold), the photoelectric current:
(a) Remains the same
(b) Doubles and KE of electrons doubles
(c) Doubles, but KE of electrons remains unchanged ✓
(d) Increases and stopping potential increases
Answer: (c)
Intensity determines the number of photons, not their energy. More photons → more photoelectrons (higher current), but KE_max = hν − φ depends only on frequency, which is unchanged. Stopping potential is also unchanged.
Intensity ↑ → saturation current ↑ (more electrons), NOT more KE. Frequency ↑ → more KE, higher stopping potential. This is the single most-tested concept in this chapter.
MCQ 2The de-Broglie wavelength of a particle is inversely proportional to its:
(a) Mass only
(b) Charge
(c) Momentum ✓
(d) Kinetic energy
Answer: (c)
λ = h/p, where p is momentum. Higher the momentum, shorter the de-Broglie wavelength. For non-relativistic particles: λ = h/√(2mKE), so λ ∝ 1/√(KE) — but the primary relation is with momentum.
λ = h/p is the primary relation. λ = h/√(2mKE) is secondary. Both appear in MCQs — always use the primary form unless kinetic energy is explicitly given.
Unit VIII · Part of 12 Marks
Atoms and Nuclei
Chapters 12–13
12
Atoms
High yieldNEET
High-yield MCQ topics
Rutherford's model & limitations
Bohr's postulates
Hydrogen spectrum series
Radius of nth orbit (r ∝ n²)
Energy of nth orbit (E ∝ −1/n²)
MCQ 1The hydrogen spectral series that lies in the visible range of the electromagnetic spectrum is:
(a) Lyman series
(b) Balmer series ✓
(c) Paschen series
(d) Brackett series
Answer: (b)
Balmer series (transitions ending at n=2) lies in the visible range (~400–700 nm). Lyman series → UV (n=1); Paschen and Brackett → infrared (n=3 and n=4 respectively).
Memorise: Lyman → UV (n=1); Balmer → Visible (n=2); Paschen → Near IR (n=3); Brackett → Mid IR (n=4); Pfund → Far IR (n=5).
MCQ 2According to Bohr's model, the radius of the nth orbit is proportional to:
(a) n
(b) 1/n
(c) n² ✓
(d) n³
Answer: (c)
r_n = a₀n²/Z, where a₀ = 0.529 Å (Bohr radius). For hydrogen (Z=1): r_n ∝ n². The second orbit has radius 4 × 0.529 Å = 2.116 Å.
r_n ∝ n² (radius grows with square of n). E_n ∝ −1/n² (energy magnitude decreases). v_n ∝ 1/n (orbital speed decreases). Know all three proportionalities.
13
Nuclei
High yieldNEET
High-yield MCQ topics
Binding energy per nucleon curve
Mass defect calculations
Nuclear fission vs fusion
Q-value of nuclear reactions
Nuclear force properties
MCQ 1Which of the following nuclei has the highest binding energy per nucleon?
(a) Hydrogen (A = 1)
(b) Uranium-235 (A = 235)
(c) Iron-56 (A = 56) ✓
(d) Helium-4 (A = 4)
Answer: (c)
Iron-56 sits at the peak of the binding energy per nucleon curve (~8.8 MeV/nucleon). Nuclei lighter than iron release energy by fusion; heavier nuclei release energy by fission — both processes move toward iron's exceptional stability.
Iron-56 is the most stable nucleus — this is why both fusion of light nuclei AND fission of heavy nuclei release energy. Understanding the BE curve is essential for this chapter.
Unit IX · Remaining Marks
Electronic Devices
Chapter 14
14
Semiconductor Electronics: Materials, Devices and Simple Circuits
Board priorityNEET
High-yield MCQ topics
p-type vs n-type semiconductors
p-n junction I-V characteristics
Forward vs reverse bias
Diode as a rectifier
Energy bands in conductors/insulators
MCQ 1In a p-type semiconductor, the majority charge carriers are:
(a) Holes ✓
(b) Electrons
(c) Protons
(d) Neutrons
Answer: (a)
p-type semiconductors are doped with trivalent impurities (e.g., Boron, Indium), creating holes as majority carriers. Electrons are minority carriers in p-type. Note: the semiconductor remains electrically neutral despite having majority carriers.
p-type: holes (majority), electrons (minority). n-type: electrons (majority), holes (minority). Having more holes does NOT mean the material is positively charged overall.
MCQ 2The output frequency of a full-wave rectifier for an input frequency f is:
(a) f/2
(b) f
(c) 2f ✓
(d) 4f
Answer: (c)
A full-wave rectifier conducts during both half-cycles, producing two output pulses per input cycle. Output frequency = 2f. A half-wave rectifier conducts only during one half-cycle, so output frequency = f.
Half-wave rectifier: f_out = f_in. Full-wave rectifier: f_out = 2f_in. This comparison is a favourite 1-mark question in board exams.
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Start free MCQs practice →Pro tips to master Class 12 Physics MCQs
| Strategy | How to apply |
|---|---|
| Formula sheet first | Physics MCQs are 60% formula-based. Build a one-page formula sheet per unit before attempting any MCQs. Priority chapters: Optics, Electrostatics, EMI, and Dual Nature. |
| Unit analysis trick | When unsure of an answer, check the units of each option. Eliminating dimensionally inconsistent options can narrow four choices down to two in seconds. |
| Chapter-by-chapter mastery | Start with high-yield chapters (Electrostatics, Optics, Dual Nature). Move to the next chapter only when you can solve 10 consecutive MCQs without error. |
| Conceptual vs numerical | CBSE boards favour conceptual MCQs (Lenz's law, dielectric effect, energy bands). NEET favours numericals. Practise both types for every chapter. |
| Assertion-reason prep | CBSE has expanded assertion-reason questions. Practise identifying: (a) whether both A and R are true, and (b) whether R correctly explains A. |
| Mistake journal | Write down every wrong answer and the exact reason you chose incorrectly. Review this before the exam — it is more valuable than re-reading the textbook. |
Frequently asked questions
Which chapters carry the most MCQs in CBSE Class 12 Physics?
Electrostatics (Chapters 1–2), Ray Optics (Chapter 9), Wave Optics (Chapter 10), Dual Nature of Radiation and Matter (Chapter 11), and Moving Charges and Magnetism (Chapter 4) are historically the highest-yield chapters. These five chapters account for the majority of MCQ marks in the 30–40% MCQ weightage of the theory paper.
How are MCQs structured in the Class 12 Physics board paper?
The CBSE paper includes standalone MCQs (1 mark each), assertion-reason pairs (1 mark each), and case-based passage questions with MCQs embedded within them. Together, these form approximately 30–40% of the 70-mark theory paper. The question paper design follows the CBSE typology: 38% remembering/understanding, 32% applying, and 30% higher-order thinking.
Are NEET-level Physics MCQs harder than CBSE board MCQs?
Yes. NEET MCQs are more numerical and require multi-concept application, while CBSE board MCQs are more conceptual and formula-application based. For students targeting both, master board-level questions first to consolidate concepts, then practise NEET-difficulty numericals with multi-step calculations.
How much time should I dedicate to Physics MCQ practice daily?
A focused 20–30 minute daily MCQ session is more effective than long irregular sessions. Solve 10–15 chapter-specific questions, review all wrong answers carefully, and maintain a mistake journal. Consistent daily practice starting at least 90 days before the board exam is the most reliable strategy for scoring 90+.
