Class 12 Physics Important Questions with Answers

Class 12 Physics Important Questions cover the main concepts, formulas, numericals and derivations from all 14 chapters in the current CBSE curriculum. They test whether students can select a physical law, apply the correct equation and present the result with suitable units.

Class 12 Physics connects electric charges, circuits, magnetism, electromagnetic induction, optics, atoms, nuclei and semiconductor devices. Its questions require conceptual understanding as well as accurate mathematical application.

Use these Class 12 Physics Important Questions for the 2026–27 examinations. Practise direct concepts first, followed by derivations and numericals. Write the applicable law, substitute quantities in SI units and state the final answer clearly.

Key Takeaways

• 14 chapters: The current course contains eight chapters in Part I and six chapters in Part II.
• Four broad areas: The chapters cover electricity, magnetism, optics and modern physics.
• Numerical method: Write the formula, convert units, substitute values and check the final unit.
• Derivation method: Begin with the governing law and show every important mathematical step.

Access Class 12 Physics Important Questions in 30 Minutes

Revise the chapters in three parts:

First 10 minutes: Electrostatics, current electricity, moving charges and magnetism

Next 10 minutes: Electromagnetic induction, alternating current, electromagnetic waves and optics

Final 10 minutes: Dual nature, atoms, nuclei and semiconductor electronics

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CBSE Class 12 Physics Chapter-Wise Important Questions

The CBSE Class 12 Physics chapter-wise important questions cover concepts, derivations and calculations from both textbook parts. Use the formulas in the tables as starting points while practising each chapter.

Physics Part I: Chapters 1–8

Part I develops the relationship between electric charges, current, magnetic fields and changing electromagnetic fields. It ends with the origin and spectrum of electromagnetic waves.

Physics Part II: Chapters 9–14

Part II covers geometrical optics, wave optics, quantum behaviour, atomic structure, nuclear physics and semiconductor devices.

Important Numerical and Conceptual Questions for Class 12 Physics

These Class 12 Physics important questions and answers cover the central method used in each chapter. Numerical values can change, but the physical principle and solving steps remain the same.

Q1. Write the electrostatic force between two point charges and explain its direction.

Answer:

According to Coulomb’s law, the magnitude of force is:

F = (1/4πε₀) × |q₁q₂|/r²

Here:

• q₁ and q₂ are the charges.
• r is the distance between them.
• ε₀ is the permittivity of free space.

The force acts along the line joining the charges. Like charges repel, while unlike charges attract.

Q2. Three capacitors C₁, C₂ and C₃ are connected in parallel. Find their equivalent capacitance and the charge on each capacitor.

Answer:

In a parallel combination, each capacitor has the same potential difference V.

The equivalent capacitance is:

Ceq = C₁ + C₂ + C₃

The charge on each capacitor is:

Q₁ = C₁V
Q₂ = C₂V
Q₃ = C₃V

The total charge is:

Q = Q₁ + Q₂ + Q₃ = CeqV

Q3. How can the temperature of a conductor be calculated from its changed resistance?

Answer:

For a limited temperature range, resistance changes approximately as:

R = R₀[1 + α(T − T₀)]

Rearranging:

T = T₀ + (R − R₀)/(αR₀)

Here:

• R₀ is the resistance at temperature T₀.
• R is the resistance at temperature T.
• α is the temperature coefficient of resistance.

All resistance values must be expressed in the same unit.

Q4. Find the magnetic force on a straight current-carrying conductor placed in a uniform magnetic field.

Answer:

The magnetic force is:

F = BIl sin θ

Here:

• B is the magnetic field.
• I is the current.
• l is the conductor’s length in the field.
• θ is the angle between current and magnetic field.

If the conductor is perpendicular to the field:

F = BIl

The direction follows Fleming’s left-hand rule.

Q5. Distinguish between diamagnetic, paramagnetic and ferromagnetic materials.

Answer:

Ferromagnetic materials may retain magnetisation after the external field is removed.

Q6. A small loop is placed inside a long solenoid. How is the induced emf found when the solenoid current changes?

Answer:

The magnetic field inside a long solenoid is:

B = μ₀nI

Magnetic flux through a loop of area A is:

ΦB = BA = μ₀nIA

Using Faraday’s law:

ε = −dΦB/dt

Therefore:

ε = −μ₀nA(dI/dt)

The negative sign represents Lenz’s law. The induced current opposes the change producing it.

Q7. Derive the resonant frequency of a series LCR circuit.

Answer:

At resonance, inductive reactance equals capacitive reactance:

XL = XC

Therefore:

ωL = 1/ωC

ω² = 1/LC

Hence:

ω₀ = 1/√LC

Since ω₀ = 2πf₀:

f₀ = 1/(2π√LC)

At resonance, the impedance becomes Z = R and the current reaches its maximum value.

Q8. The magnetic-field amplitude of an electromagnetic wave is B₀. Find its electric-field amplitude.

Answer:

For an electromagnetic wave travelling in vacuum:

E₀/B₀ = c

Therefore:

E₀ = cB₀

Here, c = 3 × 10⁸ m/s.

The electric and magnetic fields are mutually perpendicular. Both fields are also perpendicular to the direction of wave propagation.

Q9. State the lens formula and explain the Cartesian sign convention used with it.

Answer:

The thin-lens formula is:

1/f = 1/v − 1/u

Here:

• f is the focal length.
• v is the image distance.
• u is the object distance.

All distances are measured from the optical centre. Distances measured in the direction of incident light are positive, while those measured opposite are negative.

Q10. In Young’s double-slit experiment, how does fringe width depend on wavelength, screen distance and slit separation?

Answer:

The fringe width is:

β = λD/d

Here:

• λ is the wavelength.
• D is the distance between the slits and screen.
• d is the slit separation.

Fringe width increases with wavelength and screen distance. It decreases when slit separation increases.

Q11. State Einstein’s photoelectric equation and explain the threshold frequency.

Answer:

Einstein’s photoelectric equation is:

hν = ϕ + Kmax

Since Kmax = eV₀:

hν = ϕ + eV₀

The threshold frequency ν₀ is the minimum frequency needed for photoelectric emission.

ϕ = hν₀

Light below the threshold frequency cannot eject electrons, even when its intensity is increased.

Q12. Write Bohr’s expression for the energy of an electron in the nth orbit of hydrogen.

Answer:

The energy in the nth orbit is:

En = −13.6/n² eV

The negative sign shows that the electron is bound to the nucleus.

For a transition from an initial orbit ni to a final orbit nf:

hν = Ei − Ef

Emission occurs when the electron moves to a lower energy level. Absorption occurs during a transition to a higher level.

Q13. How are mass defect and nuclear binding energy calculated?

Answer:

For a nucleus containing Z protons and N neutrons:

Δm = Zmp + Nmn − Mnucleus

The binding energy is:

B.E. = Δmc²

When mass defect is measured in atomic mass units:

B.E. = Δm × 931.5 MeV

Binding energy per nucleon is:

B.E. per nucleon = Total binding energy/A

A larger value generally indicates greater nuclear stability.

Q14. Explain how a p-n junction diode works as a rectifier.

Answer:

A p-n junction diode allows significant current during forward bias and blocks most current during reverse bias.

In a half-wave rectifier, the diode conducts during one half of the alternating input cycle. It blocks the other half.

The output is pulsating direct current. A filter circuit can reduce the variations and produce a smoother output.

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