Class 12 Physics combines mathematical reasoning, conceptual comprehension, and precise formula application. Through chapter-by-chapter questions, thorough numerical solutions, and crucial derivations, NCERT Physics Class 12 Solutions (2026–27) assist students in overcoming these obstacles.
Each section explains important concepts and their applications, from comprehending electric fields and magnetic effects to solving optics problems and investigating atomic structure. This will help you solidify your foundations, practice methodically, and get ready for the CBSE Physics board exam.
Class 12 Physics Chapters in the 2026–27 CBSE Syllabus
The CBSE Class 12 Physics theory paper (subject code 042) is worth 70 marks and consists of 14 chapters divided into two sections that correspond to the NCERT textbook. Always double-check your own notes against the official CBSE curriculum PDF because topics that NCERT has removed for 2026–2027 are not evaluated.
Part | Chapters |
Part I | 1. Electric Charges and Fields · 2. Electrostatic Potential and Capacitance · 3. Current Electricity · 4. Moving Charges and Magnetism · 5. Magnetism and Matter · 6. Electromagnetic Induction · 7. Alternating Current · 8. Electromagnetic Waves |
Part II | 9. Ray Optics and Optical Instruments · 10. Wave Optics · 11. Dual Nature of Radiation and Matter · 12. Atoms · 13. Nuclei · 14. Semiconductor Electronics: Materials, Devices and Simple Circuits |
Electrostatics and Current Electricity (Chapters 1–3)
Chapter 1: Electric Charges and Fields
Q. Why can two electric field lines never cross each other?
Answer: Because the electric field at any point has only one direction. The tangent to a field line gives that direction. If two lines crossed, there would be two tangents, and so two directions of force on a test charge at the same point, which is impossible. Field lines also start on positive charges and end on negative charges; they never form closed loops in electrostatics.
Numerical. Two point charges of +2 μC and +3 μC are 30 cm apart in air. Find the force between them.
- Given: q₁ = 2 × 10⁻⁶ C, q₂ = 3 × 10⁻⁶ C, r = 0.30 m
- Formula: Coulomb’s law, F = k q₁q₂ / r². It applies to point charges at rest.
- Solution: F = (9 × 10⁹ × 2 × 10⁻⁶ × 3 × 10⁻⁶) / (0.30)² = (5.4 × 10⁻²) / (9 × 10⁻²) = 0.6 N
- Final answer: F = 0.6 N, repulsive, because both charges are positive.
The most common slip here is leaving r in centimetres. Convert before you substitute.
Derivation: field of an infinite plane sheet of charge. Take a cylindrical Gaussian surface of cross-section A, passing through the sheet with equal lengths on both sides. The field is perpendicular to the sheet, so the curved surface contributes zero flux and each flat end contributes EA. Total flux = 2EA. The enclosed charge is σA. By Gauss’s law, 2EA = σA/ε₀, so
E = σ / 2ε₀
Notice that E does not depend on distance from the sheet. Don’t confuse this with the field just outside a charged conductor, which is σ/ε₀.
Quick numerical. A closed surface encloses a net charge of 8.85 nC. The net flux is Φ = q/ε₀ = (8.85 × 10⁻⁹)/(8.85 × 10⁻¹²) = 1000 N m² C⁻¹. The shape and size of the surface do not matter, only the enclosed charge.
Chapter 2: Electrostatic Potential and Capacitance
Q. A parallel plate capacitor is charged and then disconnected from the battery. A dielectric slab is inserted. What happens to charge, capacitance, potential difference and stored energy?
Answer: The charge stays constant because the plates are isolated. Capacitance increases (C = Kε₀A/d). Since V = Q/C, the potential difference decreases. Energy U = Q²/2C also decreases. If the battery had stayed connected, V would stay constant instead, and Q and U would increase. Always decide first which quantity is held constant.
Numerical. Capacitors of 3 μF and 6 μF are connected in series across a 12 V battery. Find the charge on each, the potential difference across each and the total energy stored.
- Given: C₁ = 3 μF, C₂ = 6 μF, V = 12 V
- Formula: For series, 1/C = 1/C₁ + 1/C₂. Charge is the same on each capacitor in series.
- Solution: C = (3 × 6)/(3 + 6) = 2 μF. Q = CV = 2 × 12 = 24 μC. V₁ = Q/C₁ = 24/3 = 8 V and V₂ = Q/C₂ = 24/6 = 4 V (they add up to 12 V). U = ½CV² = ½ × 2 × 10⁻⁶ × 144 = 1.44 × 10⁻⁴ J.
- Final answer: Q = 24 μC on each; V₁ = 8 V, V₂ = 4 V; U = 144 μJ.
Check: ½QV = ½ × 24 μC × 12 V = 144 μJ. Using a second formula to verify is a good habit.
Chapter 3: Current Electricity
Q. Drift velocity is very small, yet a bulb glows almost the moment you switch it on. Why?
Answer: Switching on sets up an electric field along the whole wire at a speed close to the speed of light. Free electrons everywhere in the wire start drifting together. The bulb doesn’t wait for an electron to travel from the switch to the filament.
Numerical. A copper wire of cross-section 1.0 mm² carries 2 A. If the free-electron density is 8.5 × 10²⁸ m⁻³, find the drift velocity.
- Given: I = 2 A, A = 1.0 × 10⁻⁶ m², n = 8.5 × 10²⁸ m⁻³, e = 1.6 × 10⁻¹⁹ C
- Formula: I = n e A v_d, so v_d = I / (n e A)
- Solution: v_d = 2 / (8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.0 × 10⁻⁶) = 2 / (1.36 × 10⁴) ≈ 1.5 × 10⁻⁴ m s⁻¹
- Final answer: about 0.15 mm per second.
Numerical (terminal voltage). A cell of emf 1.5 V and internal resistance 0.5 Ω is connected to a 2.5 Ω resistor. Current I = ε/(R + r) = 1.5/3.0 = 0.5 A. Terminal voltage V = ε − Ir = 1.5 − 0.5 × 0.5 = 1.25 V, which matches IR = 0.5 × 2.5 = 1.25 V. The 0.25 V difference is the voltage lost inside the cell.
Magnetism, Induction and Alternating Current (Chapters 4–
Chapter 4: Moving Charges and Magnetism
Q. Why does magnetic field do no work on a moving charged particle?
Answer: The magnetic force F = qv × B is always perpendicular to the velocity. A force perpendicular to displacement does zero work, so the speed and kinetic energy of the particle do not change. The field only changes the direction of motion.
Numerical. A proton moves at 2 × 10⁶ m s⁻¹ perpendicular to a uniform magnetic field of 0.1 T. Find the radius of its circular path. (m = 1.67 × 10⁻²⁷ kg)
- Given: v = 2 × 10⁶ m s⁻¹, B = 0.1 T, q = 1.6 × 10⁻¹⁹ C, m = 1.67 × 10⁻²⁷ kg
- Formula: The magnetic force provides the centripetal force, so qvB = mv²/r, which gives r = mv/(qB).
- Solution: r = (1.67 × 10⁻²⁷ × 2 × 10⁶) / (1.6 × 10⁻¹⁹ × 0.1) = (3.34 × 10⁻²¹) / (1.6 × 10⁻²⁰) ≈ 0.21 m
- Final answer: r ≈ 0.21 m (21 cm).
Chapter 5: Magnetism and Matter
Q. Why can’t a magnet be broken to get an isolated north pole?
Answer: Magnetic field lines form closed loops, so the net magnetic flux through any closed surface is zero (Gauss’s law for magnetism). Cutting a bar magnet only creates two smaller magnets, each with both poles.
- Distinguish diamagnetic, paramagnetic and ferromagnetic materials.
Type | Susceptibility | Behaviour in a field | Example |
Diamagnetic | Small, negative | Weakly repelled | Bismuth, copper |
Paramagnetic | Small, positive | Weakly attracted | Aluminium |
Ferromagnetic | Large, positive | Strongly attracted | Iron, cobalt, nickel |
Ferromagnetic materials become paramagnetic above their Curie temperature.
Chapter 6: Electromagnetic Induction
Q. How does Lenz’s law follow from conservation of energy?
Answer: The induced current always opposes the change that causes it. If a magnet is pushed towards a coil, the coil’s induced current creates a pole that repels it, so you must do work to push it in. That work becomes electrical energy. If the induced current helped the motion instead, the magnet would speed up by itself while also producing current, which would create energy from nothing.
Numerical. A coil of 100 turns and area 0.02 m² is placed with its plane perpendicular to a magnetic field. The field rises uniformly from 0 to 0.5 T in 0.1 s. Find the induced emf.
- Given: N = 100, A = 0.02 m², ΔB = 0.5 T, Δt = 0.1 s
- Formula: |ε| = N ΔΦ/Δt, where Φ = BA when the field is perpendicular to the coil.
- Solution: ΔΦ = 0.5 × 0.02 = 0.01 Wb. |ε| = 100 × 0.01 / 0.1 = 10 V
- Final answer: 10 V. The negative sign in Faraday’s law only tells you the direction (Lenz’s law), so state the magnitude when asked for the value.
Chapter 7: Alternating Current
Numerical. A series LCR circuit has R = 30 Ω, X_L = 70 Ω and X_C = 30 Ω, and is connected to a 100 V (rms) AC source. Find the impedance, current, power factor and average power.
- Formula: Z = √(R² + (X_L − X_C)²), I = V/Z, cos φ = R/Z, P = V I cos φ
- Solution: Z = √(30² + 40²) = √2500 = 50 Ω. I = 100/50 = 2 A. cos φ = 30/50 = 0.6. P = 100 × 2 × 0.6 = 120 W
- Check: P = I²R = 4 × 30 = 120 W ✓
- Final answer: Z = 50 Ω, I = 2 A, power factor 0.6, average power 120 W.
Power is dissipated only in the resistor. Inductors and capacitors store and return energy, so they use no average power. At resonance, X_L = X_C, Z = R and the power factor is 1.
Chapter 8: Electromagnetic Waves
Q. Why did Maxwell introduce displacement current?
Answer: While a capacitor charges, current flows in the wires but not across the gap, so Ampere’s law gave different answers depending on which surface you chose. Maxwell added a term, displacement current I_d = ε₀ dΦ_E/dt, produced by a changing electric flux. It makes Ampere’s law consistent and leads to the prediction of electromagnetic waves.
Numerical. Find the wavelength of an electromagnetic wave of frequency 6 × 10¹⁴ Hz in vacuum. Using c = fλ, λ = (3 × 10⁸)/(6 × 10¹⁴) = 5 × 10⁻⁷ m = 500 nm, which lies in the visible region.
Optics (Chapters 9–10)
Chapter 9: Ray Optics and Optical Instruments
Numerical. An object is placed 20 cm in front of a convex lens of focal length 15 cm. Find the position, nature and magnification of the image.
- Given (new Cartesian convention): u = −20 cm, f = +15 cm
- Formula: 1/v − 1/u = 1/f, and m = v/u
- Solution: 1/v = 1/f + 1/u = 1/15 − 1/20 = (4 − 3)/60 = 1/60, so v = +60 cm. m = 60/(−20) = −3
- Final answer: The image is real, inverted, 60 cm on the other side of the lens, and three times the size of the object.
A positive v means a real image on the opposite side. Writing the sign convention at the start of your answer helps you avoid sign errors and earns method marks.
- Find the critical angle for light going from glass (n = 1.5) to air.
sin C = 1/n = 1/1.5 = 0.667, so C ≈ 42°. Total internal reflection needs two conditions: light must travel from a denser to a rarer medium, and the angle of incidence must exceed C.
Chapter 10: Wave Optics
Numerical. In Young’s double-slit experiment, the slits are 0.3 mm apart and the screen is 1.5 m away. Light of wavelength 600 nm is used. Find the fringe width. What happens if the whole setup is immersed in water (n = 4/3)?
- Formula: β = λD/d
- Solution: β = (600 × 10⁻⁹ × 1.5)/(0.3 × 10⁻³) = 3 × 10⁻³ m = 3 mm
- In water: the wavelength becomes λ/n, so the fringe width becomes β/n = 3/(4/3) = 2.25 mm. The fringes get closer together.
Q. Why don’t two separate sodium lamps produce a stable interference pattern?
Answer: Light from independent sources has phases that change randomly and extremely rapidly, so there is no constant phase difference between them. Sustained interference needs coherent sources, which is why Young’s experiment splits light from one source.
Modern Physics and Electronics (Chapters 11–14)
Chapter 11: Dual Nature of Radiation and Matter
Q. Which observations of the photoelectric effect can’t be explained by the wave theory of light?
Answer: Three observations: (1) there is a threshold frequency below which no emission occurs, however intense the light; (2) the maximum kinetic energy of photoelectrons depends on frequency, not intensity; (3) emission is almost instantaneous. Einstein’s photon picture explains all three: K_max = hν − φ.
Numerical. Light of wavelength 400 nm falls on a metal of work function 2.0 eV. Find the maximum kinetic energy, the stopping potential and the threshold wavelength.
- Formula: E = hc/λ, K_max = E − φ, eV₀ = K_max, λ₀ = hc/φ
- Solution: E = 1240/400 = 3.1 eV. K_max = 3.1 − 2.0 = 1.1 eV. V₀ = 1.1 V. λ₀ = 1240/2.0 = 620 nm
- Final answer: K_max = 1.1 eV, V₀ = 1.1 V, λ₀ = 620 nm.
Q. Find the de Broglie wavelength of an electron accelerated through 100 V. Using λ ≈ 1.227/√V nm, λ = 1.227/10 ≈ 0.123 nm, comparable to atomic spacing, which is why electrons can show diffraction from crystals.
Chapter 12: Atoms
Numerical. Find the wavelength of the photon emitted when a hydrogen atom drops from n = 3 to n = 2.
- Formula: Eₙ = −13.6/n² eV, and the photon energy is the difference between levels.
- Solution: ΔE = 13.6 × (1/2² − 1/3²) = 13.6 × (5/36) ≈ 1.89 eV. λ = 1240/1.89 ≈ 656 nm
- Final answer: about 656 nm, in the red part of the visible spectrum (the Balmer series).
Remember Bohr’s quantisation condition: the angular momentum of the electron is an integral multiple of h/2π.
Chapter 13: Nuclei
Numerical. Calculate the binding energy and binding energy per nucleon of helium-4. Given: mass of ¹H atom = 1.007825 u, mass of neutron = 1.008665 u, mass of ⁴He atom = 4.002603 u, 1 u = 931.5 MeV/c².
- Formula: Δm = Zm_H + (A − Z)m_n − M_nucleus, and BE = Δm × 931.5 MeV. Using atomic masses works because the electron masses cancel.
- Solution: Mass of constituents = 2(1.007825) + 2(1.008665) = 4.032980 u. Δm = 4.032980 − 4.002603 = 0.030377 u. BE = 0.030377 × 931.5 ≈ 28.3 MeV. Per nucleon = 28.3/4 ≈ 7.07 MeV
- Final answer: BE ≈ 28.3 MeV; about 7.1 MeV per nucleon.
Q. A sample has a half-life of 10 days. What fraction remains after 30 days?
30 days is 3 half-lives, so the fraction is (1/2)³ = 1/8.
Chapter 14: Semiconductor Electronics
Q. How is the depletion region formed in a p-n junction?
Answer: Electrons diffuse from the n-side to the p-side, and holes diffuse the other way. Near the junction they recombine, leaving behind fixed positive ions on the n-side and negative ions on the p-side. This layer has no free carriers, and its internal electric field opposes further diffusion.
Numerical. A silicon diode (forward drop 0.7 V) is connected in series with a 430 Ω resistor and a 5 V supply in forward bias. Find the current. I = (5 − 0.7)/430 = 4.3/430 = 0.010 A = 10 mA.
Common Mistakes That Cost Marks in Class 12 Physics
- Skipping unit conversion. Centimetres to metres, μF to F, nm to m. Convert before substituting.
- Ignoring sign conventions. In optics, write the convention first and keep signs with every value.
- Mixing similar formulas. For example, σ/2ε₀ for an infinite sheet versus σ/ε₀ near a conductor. Ask what the situation is before choosing the formula.
- Confusing rms and peak values in AC. Ammeters and voltmeters show rms values.
- Rounding too early. Carry extra digits and round only the final answer.
- Writing derivations without a diagram or a starting statement. State the law used, draw the figure, then derive.
How to Use NCERT Solutions for Revision
- Read the concept, then try the question yourself. Spend 10–15 minutes before looking at any solution.
- Compare, don’t copy. Check not only your answer but why that formula was chosen.
- Keep an error log. Note whether each mistake was conceptual, a formula slip or a calculation error, and revisit the log every week.
- Repeat after a gap. Redo the questions you got wrong after a few days, without the solution in front of you.
Finding Class 12 Physics Difficult? Get Help from an Experienced Physics Teacher
Written solutions show you how a problem is solved. They can’t show you where your thinking went off track. If you follow the answers but still get stuck when a numerical looks different, or you can’t recall a derivation without the book, guided practice can help. A teacher can spot the gap in a concept, correct problem-solving habits early and give you regular practice, so you prepare with more confidence. Many students need exactly this, and it’s a normal part of learning Physics.
Final Thoughts
Physics demands conceptual clarity, proper application of formulas and consistent practice to score well. NCERT Physics Class 12 Solutions (2026-27) help you revise important concepts, practice numericals and understand derivations for all 14 chapters.
Solve questions on your own, analyze your mistakes, and work on the topics that are difficult for you. Reach out to Physics Point for more help and customized learning support to make your Class 12 Physics preparation more effective.
Frequently Asked Questions
How can I prepare Class 12 Physics using NCERT solutions?
Study a chapter’s concepts, attempt the in-text and exercise questions on your own, and then use the solutions to check your method. Repeat the questions you got wrong after a few days.
Are NCERT solutions enough for CBSE Class 12 Physics board preparation?
The CBSE syllabus is built on the NCERT textbook, so it is the essential base: concepts, derivations, solved examples and exercises. Add CBSE sample papers and previous papers to practise the paper pattern and time management. No single resource can guarantee a particular question or score.
How should I practise Class 12 Physics numericals?
Write the given data with units, name the principle, choose the formula, then solve step by step and check the unit and the order of magnitude. After solving, try the same problem with changed numbers.
Which Class 12 Physics chapters should I revise first?
Start with chapters that other chapters depend on: Electric Charges and Fields, Capacitance and Current Electricity. Then plan by the unit-wise marks in the official CBSE syllabus, giving more time to the units where you are weakest.
How can I improve my understanding of Physics derivations?
Learn the starting law and the key idea first, for example Gauss’s law with a suitable surface, then practise writing the derivation from memory with a diagram. Don’t memorise it line by line.
How often should I practise NCERT Physics questions?
Short, regular sessions work better than occasional long ones. Aim for some numericals most days and a weekly review of your error log.
