Let’s be honest: Physics is the subject that either clicks for students or keeps them up at night. If you’re aiming for engineering, medicine, technology or another science-based course, a strong Class 12 Physics score really matters. And getting there takes more than memorising definitions and formulas. You need to understand the concepts, practise numericals regularly, get comfortable with the key derivations, and learn to apply Physics ideas to real-life situations. Still thinking about where Physics might take you? Our guide to the best career options after Class 12 is a good place to start.
With the board exams getting closer, the smartest thing you can do is stay close to the prescribed syllabus, practise previous-year papers, and work through questions that match the latest exam pattern.
To make your revision easier, we’ve put together 30 important CBSE Class 12 Physics questions and answers covering all nine units of the 2026–27 syllabus. You’ll find conceptual questions, numericals, derivations, application-based questions and a case study, so you get a feel for the different kinds of questions you may meet in the paper.
Whether you’re revising Electrostatics, Current Electricity, Optics, Modern Physics or Semiconductor Electronics, these questions are meant to deepen your understanding and help you prepare in a steady, organised way.
CBSE Class 12 Physics Exam Pattern 2026–27
Before you jump into the questions, it helps to know what the paper looks like and how the marks are spread out.
The Class 12 Physics assessment is out of 100 marks: a 70-mark theory paper and a 30-mark practical assessment.
The theory paper is three hours long.
Unit-Wise Marks Distribution
For 2026–27, CBSE groups the Physics chapters into nine units. Here is how the marks are distributed:
| Unit | Chapters / Topics | Marks |
| I | Electrostatics | |
| II | Current Electricity | 16 (combined) |
| III | Magnetic Effects of Current and Magnetism | |
| IV | Electromagnetic Induction and Alternating Currents | 17 (combined) |
| V | Electromagnetic Waves | |
| VI | Optics | 18 (combined) |
| VII | Dual Nature of Radiation and Matter | |
| VIII | Atoms and Nuclei | 12 (combined) |
| IX | Electronic Devices | 7 |
| Total | 70 |
One thing to keep in mind: the marks in this table apply to the grouped units, not to individual chapters. CBSE does not prescribe a fixed chapter-wise split in this table.
So give a little extra time to Electrostatics, Current Electricity, Magnetism, Electromagnetic Induction, Alternating Current and Optics, but don’t skip any unit. Every prescribed unit deserves a proper revision.
CBSE Class 12 Physics Question Paper Structure
The 2026–27 Physics theory paper has five sections. Here is the structure at a glance:
| Section | Question Type | Number of Questions | Marks |
| A | MCQs and Assertion-Reason | 16 | 16 |
| B | Very Short Answer | 5 | 10 |
| C | Short Answer | 7 | 21 |
| D | Case Study-Based | 2 | 8 |
| E | Long Answer | 3 | 15 |
| Total | 33 | 70 |
Try to practise every one of these formats, including MCQs, assertion-reason questions, numericals, derivations and case study-based questions, so that nothing in the paper feels unfamiliar on exam day.
CBSE Class 12 Physics Important Questions and Answers
The 30 questions below cover the major concepts from all nine units of the CBSE Class 12 Physics syllabus. Try each question on your own first, and only then check the solution. That is where the real learning happens.
Download all 30 questions and answers in a printable PDF file
Unit 1: Electrostatics
Question 1. State Coulomb’s law and calculate the electrostatic force between two point charges of 2 μC and 3 μC separated by a distance of 0.3 m in vacuum.
Answer:
Coulomb’s Law:
Coulomb’s law states that the electrostatic force between two stationary point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.
Mathematically,
F = (1/4πε₀) × |q₁q₂|/r²
Where:
- q₁ and q₂ are the magnitudes of the charges.
- r is the distance between the charges.
- ε₀ is the permittivity of free space.
- 1/4πε₀ = 9 × 10⁹ N m² C⁻².
Given:
q₁ = 2 μC = 2 × 10⁻⁶ C
q₂ = 3 μC = 3 × 10⁻⁶ C
r = 0.3 m
Substituting:
F = (9 × 10⁹ × 2 × 10⁻⁶ × 3 × 10⁻⁶)/(0.3)²
F = (54 × 10⁻³)/0.09
F = 0.6 N
Final Answer: The magnitude of the electrostatic force is 0.6 N.
Since both charges are positive, the force is repulsive.
Question 2. State Gauss’s theorem and write its mathematical expression. How is it used to calculate the electric field due to an infinitely long uniformly charged straight wire?
Answer:
Gauss’s Theorem:
Gauss’s theorem states that the total electric flux through any closed surface is equal to 1/ε₀ times the total electric charge enclosed within that surface.
Mathematically,
∮ E · dA = Qₑₙ꜀/ε₀
Where:
- E is the electric field.
- dA is the area vector.
- Qₑₙ꜀ is the charge enclosed by the Gaussian surface.
- ε₀ is the permittivity of free space.
Application to an infinitely long charged wire:
Consider an infinitely long straight wire having uniform linear charge density λ.
Choose a cylindrical Gaussian surface of radius r and length l around the wire.
Due to cylindrical symmetry, the electric field has the same magnitude at every point on the curved surface and is directed radially outward.
The electric flux through the two flat ends is zero because the electric field is perpendicular to their area vectors.
Therefore, the total electric flux is:
Φ = E × 2πrl
The charge enclosed is:
Qₑₙ꜀ = λl
Using Gauss’s theorem:
E × 2πrl = λl/ε₀
Therefore,
E = λ/(2πε₀r)
This is the electric field due to an infinitely long uniformly charged straight wire.
Question 3. Derive the expression for the capacitance of a parallel plate capacitor in vacuum. What happens when a dielectric medium is introduced between the plates?
Answer:
Consider a parallel plate capacitor consisting of two conducting plates, each having area A and separated by a distance d.
Let the charges on the plates be +Q and −Q.
Surface charge density is:
σ = Q/A
The electric field between the plates, neglecting edge effects, is:
E = σ/ε₀
Therefore,
E = Q/(Aε₀)
The potential difference between the plates is:
V = Ed
Substituting:
V = Qd/(Aε₀)
Capacitance is defined as:
C = Q/V
Therefore,
C = Q/[Qd/(Aε₀)]
C = ε₀A/d
This is the capacitance of a parallel plate capacitor in vacuum.
When a dielectric is introduced:
If a dielectric of relative permittivity K completely fills the space between the plates, the capacitance becomes:
C = Kε₀A/d
Thus, the capacitance increases by a factor of K.
Unit 2: Current Electricity
Question 4. State Ohm’s law. A conductor has a resistance of 10 Ω and carries a current of 2 A. Calculate the potential difference across it and the electrical power consumed.
Answer:
Ohm’s Law:
Ohm’s law states that, at constant temperature and other physical conditions, the current flowing through a conductor is directly proportional to the potential difference across its ends.
V ∝ I
Therefore,
V = IR
Where:
- V = Potential difference
- I = Current
- R = Resistance
Given:
R = 10 Ω
I = 2 A
Potential difference:
V = IR
= 2 × 10
= 20 V
Electrical power:
P = VI
= 20 × 2
= 40 W
Final Answer:
Potential difference = 20 V
Electrical power = 40 W
Question 5. Derive the expression for the equivalent resistance of three resistors connected in parallel.
Answer:
Consider three resistors R₁, R₂, and R₃ connected in parallel across a battery.
In a parallel combination, the potential difference across each resistor remains the same.
Let the common potential difference be V.
The currents through the resistors are:
I₁ = V/R₁
I₂ = V/R₂
I₃ = V/R₃
According to Kirchhoff’s current law, the total current is:
I = I₁ + I₂ + I₃
Substituting:
I = V/R₁ + V/R₂ + V/R₃
If R is the equivalent resistance, then:
I = V/R
Therefore,
V/R = V/R₁ + V/R₂ + V/R₃
Dividing by V:
1/R = 1/R₁ + 1/R₂ + 1/R₃
This is the expression for the equivalent resistance of three resistors connected in parallel.
Question 6. A cell of emf 12 V and internal resistance 1 Ω is connected to an external resistance of 5 Ω. Calculate the current and terminal potential difference.
Answer:
Given:
EMF (E) = 12 V
Internal resistance (r) = 1 Ω
External resistance (R) = 5 Ω
The current flowing through the circuit is:
I = E/(R + r)
Substituting:
I = 12/(5 + 1)
I = 12/6
I = 2 A
The terminal potential difference is:
V = E − Ir
V = 12 − (2 × 1)
V = 10 V
Alternatively:
V = IR = 2 × 5 = 10 V
Final Answer:
Current = 2 A
Terminal potential difference = 10 V
Unit 3: Magnetic Effects of Current and Magnetism
Question 7. State the Biot-Savart law and write the expression for the magnetic field at the centre of a circular current-carrying loop.
Answer:
Biot-Savart Law:
The Biot-Savart law states that the magnetic field produced at a point by a small current element is directly proportional to the current, the length of the current element, and the sine of the angle between the current element and the position vector. It is inversely proportional to the square of the distance.
Mathematically,
dB = (μ₀/4π) × (I dl sin θ/r²)
Where:
- dB = Small magnetic field contribution
- μ₀ = Permeability of free space
- I = Current
- dl = Current element
- r = Distance from the element
- θ = Angle between dl and r
Magnetic field at the centre of a circular loop:
For a circular loop of radius R carrying current I, every current element produces a magnetic field in the same direction at the centre.
Integrating the contributions of all elements:
B = (μ₀I/4πR²) × 2πR
Therefore,
B = μ₀I/2R
For a circular coil having N turns:
B = μ₀NI/2R
Question 8. A charged particle enters a uniform magnetic field perpendicular to the field direction. Derive the expression for the radius of its circular path.
Answer:
When a charged particle enters a uniform magnetic field perpendicular to its velocity, it experiences a magnetic Lorentz force.
The magnitude of this force is:
F = |q|vB
Where:
- q = Charge of the particle
- v = Velocity
- B = Magnetic field
This magnetic force acts perpendicular to the velocity and provides the centripetal force required for circular motion.
Therefore,
|q|vB = mv²/r
Where m is the mass of the particle and r is the radius of the circular path.
Rearranging:
r = mv²/(|q|vB)
r = mv/(|q|B)
Thus, the radius of the circular path is directly proportional to the particle’s mass and velocity and inversely proportional to the magnetic field and magnitude of charge.
Question 9. Explain the principle and working of a moving coil galvanometer.
Answer:
Principle:
A current-carrying coil placed in a magnetic field experiences a torque.
The torque acting on the coil is:
τ = NIAB sin θ
Where:
- N = Number of turns
- I = Current
- A = Area of the coil
- B = Magnetic field
- θ = Angle between the normal to the coil and the magnetic field.
In a radial magnetic field, the deflecting torque is:
τ = NIAB
The restoring torque produced by the suspension is:
τ = kφ
Where k is the torsional constant and φ is the angular deflection.
At equilibrium:
NIAB = kφ
Therefore,
I = kφ/NAB
Working:
A rectangular coil is suspended between the poles of a permanent magnet. When current flows through the coil, it experiences a torque and rotates.
The suspension produces an opposing restoring torque. The coil comes to rest when both torques become equal.
The deflection is directly proportional to the current.
Hence, a moving coil galvanometer can be used to detect and measure small electric currents.
Unit 4: Electromagnetic Induction and Alternating Currents
Question 10. State Faraday’s laws of electromagnetic induction and explain Lenz’s law.
Answer:
Faraday’s First Law:
Whenever the magnetic flux linked with a closed circuit changes, an electromotive force is induced in the circuit.
If the circuit is closed, an induced current flows through it.
Faraday’s Second Law:
The magnitude of the induced electromotive force is directly proportional to the rate of change of magnetic flux linked with the circuit.
For a coil of N turns:
E = −N(dΦB/dt)
Where:
- E = Induced emf
- N = Number of turns
- ΦB = Magnetic flux
Lenz’s Law:
Lenz’s law states that the direction of induced current is such that the magnetic field produced by it opposes the change in magnetic flux responsible for its induction.
The negative sign in Faraday’s law represents Lenz’s law.
This law is consistent with the principle of conservation of energy.
Question 11. A coil of 200 turns experiences a change in magnetic flux from 5 × 10⁻³ Wb to 1 × 10⁻³ Wb in 0.02 seconds. Calculate the magnitude of the induced emf.
Answer:
Given:
Number of turns, N = 200
Initial flux = 5 × 10⁻³ Wb
Final flux = 1 × 10⁻³ Wb
Time interval = 0.02 s
Change in flux:
ΔΦ = (1 × 10⁻³) − (5 × 10⁻³)
ΔΦ = −4 × 10⁻³ Wb
Magnitude of change:
|ΔΦ| = 4 × 10⁻³ Wb
Using Faraday’s law:
|E| = N × |ΔΦ|/Δt
= 200 × (4 × 10⁻³)/0.02
= 0.8/0.02
= 40 V
Final Answer: The magnitude of the induced emf is 40 V.
Question 12. Derive the expression for the impedance of a series LCR circuit connected to an alternating voltage source.
Answer:
Consider a resistor R, an inductor L, and a capacitor C connected in series to an alternating voltage source.
The voltage across the resistor is in phase with the current.
The voltage across the inductor leads the current by 90°.
The voltage across the capacitor lags the current by 90°.
The inductive reactance is:
Xᴸ = ωL
The capacitive reactance is:
Xᶜ = 1/ωC
The net reactive voltage is determined by the difference between the inductive and capacitive voltage drops.
Using the phasor diagram:
V² = Vᴿ² + (Vᴸ − Vᶜ)²
Dividing by I²:
(V/I)² = R² + (Xᴸ − Xᶜ)²
Since impedance Z = V/I:
Z = √[R² + (Xᴸ − Xᶜ)²]
Therefore,
Z = √[R² + (ωL − 1/ωC)²]
This is the impedance of a series LCR circuit.
At resonance:
Xᴸ = Xᶜ
Therefore, impedance becomes minimum and is equal to R.
The resonant angular frequency is:
ω₀ = 1/√LC
Unit 5: Electromagnetic Waves
Question 13. What are electromagnetic waves? Write any four important properties of electromagnetic waves.
Answer:
Electromagnetic waves are waves consisting of mutually perpendicular, time-varying electric and magnetic fields that propagate through space.
They are produced by accelerated electric charges.
Important properties (any four are sufficient for the answer):
- Electromagnetic waves are transverse waves.
- The electric field and magnetic field are perpendicular to each other and to the direction of propagation.
- They do not require a material medium and can travel through vacuum.
- In vacuum, they travel with a speed of approximately 3 × 10⁸ m/s.
- The electric and magnetic fields are in the same phase.
- They carry energy and momentum.
The speed of electromagnetic waves in vacuum is:
c = 1/√(μ₀ε₀)
Question 14. Arrange the following electromagnetic waves in increasing order of frequency: Radio waves, X-rays, Infrared rays, Ultraviolet rays, Microwaves and Gamma rays.
Answer:
The electromagnetic spectrum in increasing order of frequency is:
- Radio waves
- Microwaves
- Infrared rays
- Visible light
- Ultraviolet rays
- X-rays
- Gamma rays
Final Answer:
Radio waves < Microwaves < Infrared rays < Ultraviolet rays < X-rays < Gamma rays.
Visible light lies between infrared and ultraviolet radiation.
Unit 6: Optics
Question 15. State the laws of refraction. Write the lens formula and calculate the image distance for a convex lens of focal length 20 cm when an object is placed 30 cm in front of it.
Answer:
Laws of Refraction:
- The incident ray, refracted ray, and normal at the point of incidence lie in the same plane.
- For a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant.
According to Snell’s law:
n₁ sin i = n₂ sin r
Lens Formula:
1/f = 1/v − 1/u
Where:
- f = Focal length
- v = Image distance
- u = Object distance
Given:
Focal length, f = +20 cm
Object distance, u = −30 cm
Using the lens formula:
1/20 = 1/v − 1/(−30)
1/20 = 1/v + 1/30
1/v = 1/20 − 1/30
1/v = (3 − 2)/60
1/v = 1/60
v = 60 cm
Final Answer: The image is formed 60 cm on the other side of the lens.
The image is real, inverted, and magnified.
Question 16. What is total internal reflection? State the necessary conditions for total internal reflection.
Answer:
Total Internal Reflection:
Total internal reflection is the phenomenon in which a light ray travelling from an optically denser medium to an optically rarer medium is completely reflected back into the denser medium when the angle of incidence exceeds the critical angle.
Necessary conditions:
- Light must travel from an optically denser medium to an optically rarer medium.
- The angle of incidence must be greater than the critical angle for the two media.
Applications:
- Optical fibres
- Endoscopy
- Reflecting prisms
- Brilliance of diamonds
Question 17. In Young’s double-slit experiment, the distance between two slits is 0.5 mm and the screen is placed 2 m away. If the wavelength of light is 500 nm, calculate the fringe width.
Answer:
The fringe width in Young’s double-slit experiment is:
β = λD/d
Where:
- λ = Wavelength of light
- D = Distance between the slits and screen
- d = Separation between the slits
Given:
λ = 500 nm = 500 × 10⁻⁹ m
D = 2 m
d = 0.5 mm = 0.5 × 10⁻³ m
Substituting:
β = (500 × 10⁻⁹ × 2)/(0.5 × 10⁻³)
β = (1000 × 10⁻⁹)/(0.5 × 10⁻³)
β = 2 × 10⁻³ m
Converting into millimetres:
β = 2 mm
Final Answer: The fringe width is 2 mm.
Question 18. Explain the phenomenon of diffraction of light. How does it differ from interference?
Answer:
Diffraction:
Diffraction is the phenomenon in which light waves bend and spread around the edges of an obstacle or through a narrow opening when the dimensions are comparable to the wavelength of light.
It demonstrates the wave nature of light.
Difference between Diffraction and Interference:
| Interference | Diffraction |
| It results from the superposition of waves originating from coherent sources. | It results from the superposition of wavelets from different parts of the same wavefront. |
| Fringes may have nearly equal widths in Young’s double-slit experiment. | The central maximum in single-slit diffraction is wider than the secondary maxima. |
| Bright fringes can have approximately equal intensity in an ideal arrangement of equal sources. | Secondary maxima have decreasing intensity. |
Both phenomena demonstrate the wave nature of light.
Unit 7: Dual Nature of Radiation and Matter
Question 19. Explain the photoelectric effect and state Einstein’s photoelectric equation.
Answer:
Photoelectric Effect:
The photoelectric effect is the phenomenon in which electrons are emitted from the surface of a metal when electromagnetic radiation of sufficiently high frequency falls on it.
The emitted electrons are called photoelectrons.
Einstein’s Photoelectric Equation:
According to Einstein, the energy of an incident photon is used partly to overcome the work function of the metal and the remaining energy appears as the maximum kinetic energy of the emitted electron.
Therefore,
hν = φ + Kₘₐₓ
Where:
- h = Planck’s constant
- ν = Frequency of incident radiation
- φ = Work function of the metal
- Kₘₐₓ = Maximum kinetic energy of the photoelectron
Since Kₘₐₓ = eV₀:
hν = φ + eV₀
Where V₀ is the stopping potential.
The threshold frequency is:
ν₀ = φ/h
The maximum kinetic energy depends on the frequency of incident radiation, not on its intensity.
Question 20. Light of frequency 8 × 10¹⁴ Hz falls on a metal having a work function of 2 eV. Calculate the maximum kinetic energy of the emitted photoelectrons.
Answer:
Given:
Frequency, ν = 8 × 10¹⁴ Hz
Work function, φ = 2 eV
Planck’s constant, h = 6.63 × 10⁻³⁴ Js
Energy of the incident photon:
E = hν
= 6.63 × 10⁻³⁴ × 8 × 10¹⁴
= 5.304 × 10⁻¹⁹ J
Converting into electron volts:
E = (5.304 × 10⁻¹⁹)/(1.602 × 10⁻¹⁹)
≈ 3.31 eV
Using Einstein’s equation:
Kₘₐₓ = hν − φ
= 3.31 − 2
= 1.31 eV
Final Answer: The maximum kinetic energy is approximately 1.31 eV.
Unit 8: Atoms and Nuclei
Question 21. State the main postulates of Bohr’s atomic model.
Answer:
Niels Bohr proposed an atomic model to explain the stability of atoms and the discrete spectral lines of hydrogen.
The main postulates are:
- Electrons revolve around the nucleus in certain permitted circular orbits called stationary orbits.
- Electrons do not radiate energy while moving in these permitted stationary orbits.
- The angular momentum of an electron in an allowed orbit is quantised.
mvr = nh/2π
Where n = 1, 2, 3, …
- Radiation is emitted or absorbed when an electron jumps from one permitted energy level to another.
The frequency of the emitted or absorbed radiation is determined by:
hν = |E₂ − E₁|
These postulates explain the discrete energy levels of the hydrogen atom.
Question 22. Calculate the energy of a photon emitted when an electron in a hydrogen atom jumps from the n = 3 energy level to the n = 2 energy level.
Answer:
The energy of an electron in the nth orbit of a hydrogen atom is:
Eₙ = −13.6/n² eV
For n = 3:
E₃ = −13.6/9
= −1.51 eV (approximately)
For n = 2:
E₂ = −13.6/4
= −3.4 eV
The energy of the emitted photon is:
ΔE = E₃ − E₂ in magnitude
= |−1.51 − (−3.4)|
= 1.89 eV
Final Answer: The energy of the emitted photon is approximately 1.89 eV.
Question 23. Define mass defect and binding energy. Calculate the binding energy of a nucleus if its mass defect is 0.02 u.
Answer:
Mass Defect:
Mass defect is the difference between the sum of the masses of the individual nucleons and the actual mass of the nucleus.
Δm = Zmₚ + (A − Z)mₙ − M
Where:
- Z = Number of protons
- A − Z = Number of neutrons
- mₚ = Mass of a proton
- mₙ = Mass of a neutron
- M = Actual nuclear mass
Binding Energy:
Binding energy is the energy required to separate a nucleus completely into its constituent protons and neutrons.
According to Einstein’s mass-energy relation:
E = Δmc²
Given:
Δm = 0.02 u
We know:
1 u = 931.5 MeV/c²
Therefore:
Binding energy = 0.02 × 931.5 MeV
= 18.63 MeV
Final Answer: The binding energy of the nucleus is 18.63 MeV.
Question 24. Differentiate between nuclear fission and nuclear fusion.
Answer:
| Nuclear Fission | Nuclear Fusion |
| A heavy nucleus splits into two or more lighter nuclei. | Two light nuclei combine to form a heavier nucleus. |
| It generally produces neutrons and releases energy. | It releases a large amount of energy. |
| It can be initiated by neutron bombardment in suitable fissile materials. | It requires extremely high temperatures and suitable conditions. |
| It is used in nuclear reactors. | It is the main source of energy in the Sun and other stars. |
Both processes release energy due to changes in nuclear binding energy.
Unit 9: Electronic Devices
Question 25. What is a semiconductor? Explain the difference between intrinsic and extrinsic semiconductors.
Answer:
A semiconductor is a material whose electrical conductivity lies between that of a conductor and an insulator.
Its conductivity can be controlled by temperature, impurities, and other external conditions.
Intrinsic Semiconductor:
An intrinsic semiconductor is a pure semiconductor without intentionally added impurities.
Examples include pure silicon and germanium.
In an intrinsic semiconductor:
- The number of free electrons equals the number of holes.
- Its electrical conductivity is relatively low at room temperature.
Extrinsic Semiconductor:
An extrinsic semiconductor is formed by adding a small amount of suitable impurity to a pure semiconductor. This process is called doping.
There are two types:
1. n-type Semiconductor:
It is formed by doping silicon or germanium with a pentavalent impurity such as phosphorus.
Electrons are the majority charge carriers, while holes are the minority carriers.
2. p-type Semiconductor:
It is formed by doping silicon or germanium with a trivalent impurity such as boron.
Holes are the majority charge carriers, while electrons are the minority carriers.
Question 26. Explain the formation of a p-n junction diode and describe its working under forward and reverse bias.
Answer:
A p-n junction diode is formed by joining p-type and n-type semiconductor materials.
When the two regions are joined, electrons from the n-region diffuse towards the p-region, while holes from the p-region diffuse towards the n-region.
This creates a depletion region near the junction and an internal potential barrier.
Forward Bias:
In forward bias:
- The p-side is connected to the positive terminal of the battery.
- The n-side is connected to the negative terminal.
- The depletion layer becomes thinner.
- The potential barrier decreases.
- A substantial current flows after the applied voltage becomes sufficient.
Reverse Bias:
In reverse bias:
- The p-side is connected to the negative terminal.
- The n-side is connected to the positive terminal.
- The depletion layer becomes wider.
- The potential barrier increases.
- Only a very small reverse saturation current flows under normal operating conditions.
Conclusion:
A p-n junction diode allows current to flow much more easily in the forward direction than in the reverse direction. Therefore, it is used as a rectifier.
Additional Important Numerical and Conceptual Questions
The questions below give you some extra practice with numerical problem-solving and with applying Physics concepts.
Question 27. Two resistors of 4 Ω and 6 Ω are connected in series to a 20 V battery. Calculate the current flowing through the circuit and the potential difference across each resistor.
Answer:
Given:
R₁ = 4 Ω
R₂ = 6 Ω
V = 20 V
For resistors in series:
R = R₁ + R₂
R = 4 + 6
R = 10 Ω
Using Ohm’s law:
I = V/R
= 20/10
= 2 A
Potential difference across the first resistor:
V₁ = IR₁
= 2 × 4
= 8 V
Potential difference across the second resistor:
V₂ = IR₂
= 2 × 6
= 12 V
Final Answer:
Current = 2 A
Potential difference across 4 Ω resistor = 8 V
Potential difference across 6 Ω resistor = 12 V
Question 28. A transformer has 500 turns in its primary coil and 100 turns in its secondary coil. If the primary voltage is 220 V, calculate the secondary voltage. Also identify the type of transformer.
Answer:
Given:
Primary turns, Nₚ = 500
Secondary turns, Nₛ = 100
Primary voltage, Vₚ = 220 V
The transformer equation is:
Vₛ/Vₚ = Nₛ/Nₚ
Therefore:
Vₛ = Vₚ × Nₛ/Nₚ
= 220 × 100/500
= 220 × 0.2
= 44 V
Since the secondary voltage is less than the primary voltage, it is a step-down transformer.
Final Answer: Secondary voltage = 44 V. It is a step-down transformer.
Question 29. A convex mirror has a focal length of 20 cm. An object is placed 30 cm in front of it. Calculate the image distance.
Answer:
Given:
Focal length, f = +20 cm
Object distance, u = −30 cm
Using the mirror formula:
1/f = 1/v + 1/u
Substituting:
1/20 = 1/v − 1/30
1/v = 1/20 + 1/30
1/v = (3 + 2)/60
1/v = 5/60
1/v = 1/12
v = 12 cm
Final Answer: The image is formed 12 cm behind the mirror.
The image is virtual, erect, and diminished.
Question 30. Case Study: Photoelectric Effect
A student performs an experiment to study the photoelectric effect using a metal surface. Light of different frequencies is allowed to fall on the metal, and the stopping potential is measured.
Answer the following questions:
(a) What is the minimum frequency of incident light required to produce photoelectric emission called?
Answer:
It is called the threshold frequency (ν₀).
(b) What happens to the maximum kinetic energy of photoelectrons when the frequency of incident light increases above the threshold frequency?
Answer:
The maximum kinetic energy increases linearly with the frequency of incident light, according to Einstein’s photoelectric equation.
Kₘₐₓ = hν − φ
(c) What happens to the photoelectric current when the intensity of incident light increases, keeping its frequency above the threshold frequency?
Answer:
The photoelectric current increases because a greater number of photons fall on the metal surface per unit time, resulting in the emission of more photoelectrons.
Important Physics Formulas for CBSE Class 12 Board Exams
It’s worth revising formulas regularly, because they turn up again and again in numericals. Here is a quick reference of the essential equations from the Class 12 Physics syllabus.
1. Electrostatics
- Coulomb’s law: F = (1/4πε₀) × |q₁q₂|/r²
- Electric field: E = F/q
- Electric field due to a point charge: E = (1/4πε₀) × q/r²
- Electric potential due to a point charge: V = (1/4πε₀) × q/r
- Capacitance: C = Q/V
- Parallel plate capacitor: C = ε₀A/d
- Energy stored in a capacitor: U = ½CV²
2. Current Electricity
- Ohm’s law: V = IR
- Resistance: R = ρl/A
- Electric power: P = VI = I²R = V²/R
- Current: I = Q/t
- Cell current: I = E/(R + r)
- Drift current: I = neAvd
3. Magnetic Effects of Current and Magnetism
- Lorentz force: F = qvB sin θ
- Force on a current-carrying conductor: F = BIl sin θ
- Magnetic field at the centre of a circular coil: B = μ₀NI/2R
- Magnetic field inside a long solenoid: B = μ₀nI
- Radius of circular motion of a charged particle: r = mv/(|q|B)
4. Electromagnetic Induction and Alternating Currents
- Magnetic flux: Φ = BA cos θ
- Faraday’s law: E = −N(dΦ/dt)
- Inductive reactance: Xᴸ = ωL
- Capacitive reactance: Xᶜ = 1/ωC
- Impedance of a series LCR circuit: Z = √[R² + (Xᴸ − Xᶜ)²]
- Resonant angular frequency: ω₀ = 1/√LC
- RMS voltage: Vᵣₘₛ = V₀/√2
- RMS current: Iᵣₘₛ = I₀/√2
5. Electromagnetic Waves
- Speed of electromagnetic waves: c = 1/√(μ₀ε₀)
- Wave relation: c = νλ
- Electric and magnetic field relationship: E₀ = cB₀
6. Optics
- Mirror formula: 1/f = 1/v + 1/u
- Lens formula: 1/f = 1/v − 1/u
- Magnification for a lens: m = v/u
- Power of a lens: P = 1/f (f in metres)
- Refractive index: n = c/v
- Fringe width in Young’s double-slit experiment: β = λD/d
7. Dual Nature of Radiation and Matter
- Photon energy: E = hν
- Einstein’s photoelectric equation: hν = φ + Kₘₐₓ
- Maximum kinetic energy: Kₘₐₓ = eV₀
- de Broglie wavelength: λ = h/p
8. Atoms and Nuclei
- Energy of an electron in the nth orbit of hydrogen: Eₙ = −13.6/n² eV
- Mass-energy relation: E = Δmc²
- Binding energy: B.E. = Δmc²
- Radioactive decay law: N = N₀e⁻λᵗ
- Half-life: T₁/₂ = 0.693/λ
9. Electronic Devices
- Current gain in common-emitter configuration: β = I꜀/Iᵦ
- Current relation: Iₑ = I꜀ + Iᵦ
A quick reminder: always use the sign conventions and conditions that go with each formula. The mirror and lens formulas, in particular, must be used with the prescribed Cartesian sign convention.
How to Score 95+ Marks in CBSE Class 12 Physics
Scoring 95 or more in Physics doesn’t come from one magic trick. It comes from understanding concepts, getting your numericals right, revising regularly and using smart exam techniques. For the wider picture, you can also read our guide on how to score 95%+ in board exams.
Here are nine strategies that can help you get there.
1. Understand the Fundamental Concepts
Physics is built on a few fundamental principles and their applications. Take the time to understand what each physical quantity, law and equation actually means, instead of simply memorising formulas. When you understand the idea, the formula becomes much easier to remember and to use. Methods like active recall and spaced revision work well for Physics too, and you will find them in our guide to effective study techniques for students.
Give special attention to Electrostatics, Current Electricity, Magnetism, Electromagnetic Induction and Optics.
2. Practise NCERT Textbook Questions
The prescribed NCERT Physics textbook should be the foundation of your board exam preparation.
- Read each chapter carefully.
- Make sure you understand every definition and law.
- Solve the in-text questions and the prescribed exercises.
- Practise the derivations included in the syllabus.
- Study the diagrams and graphs closely.
3. Master Important Derivations
Derivations are a big part of Physics preparation, so practise writing them in a clear, logical sequence. A neat, step-by-step derivation is easier to follow and easier to mark.
Pay special attention to:
- Electric field due to a point charge and applications of Gauss’s law.
- Capacitance of a parallel plate capacitor.
- Equivalent resistance in series and parallel combinations.
- Magnetic field due to a current-carrying circular loop and solenoid.
- Radius of the circular path of a charged particle in a magnetic field.
- Impedance of a series LCR circuit.
- Lens formula and related optical concepts prescribed in the syllabus.
- Fringe width in Young’s double-slit experiment.
- Einstein’s photoelectric equation.
Before you spend time on extra derivations, check the latest syllabus to make sure they are actually prescribed.
4. Practise Numerical Problems Daily
Numericals need regular practice. Get into the habit of choosing the right formula, converting everything into SI units, substituting values carefully, and writing the final answer with the correct units.
Try problems of different difficulty levels from every chapter, so that a small twist in a question doesn’t catch you off guard.
5. Revise Important Diagrams and Graphs
A neat, correctly labelled diagram can make your answer much clearer.
Practise drawing:
- Electric field lines.
- Electric dipole configurations.
- Magnetic field patterns.
- AC circuit diagrams.
- Ray diagrams for lenses and mirrors.
- Young’s double-slit experiment.
- Photoelectric effect graphs.
- p-n junction diode characteristics.
6. Solve Previous-Year Question Papers
Previous-year papers help you get comfortable with the kinds of questions asked in board exams. If you’re not sure how to make the most of them, here is our guide on how to solve previous year question papers effectively.
Aim to solve at least five to ten papers under proper examination conditions.
After each paper, check your answers against reliable marking schemes and note down the areas that need more work.
7. Practise Case Study and Competency-Based Questions
CBSE tests whether you can apply scientific concepts to situations you haven’t seen before, so case study and competency-based questions deserve a regular place in your practice.
Practise questions based on:
- Electric circuits and household electricity.
- Magnetic fields and electromagnetic induction.
- Optical instruments and image formation.
- Photoelectric experiments.
- Semiconductor devices and electronic circuits.
8. Prepare a Separate Formula Notebook
Keep one notebook for important formulas, definitions, units, constants and derivations.
Dip into it regularly instead of waiting for the last few days before the exam. Short, frequent revision beats a last-minute rush.
9. Take Timed Mock Tests
Try full-length practice papers within the three-hour time limit.
During each test, practise dividing your time sensibly between MCQs, numericals, derivations and long-answer questions.
After every test, go through your mistakes carefully and make a note not to repeat them next time.
Frequently Asked Questions (FAQs)
1. Which units are most important for CBSE Class 12 Physics Board Exams?
According to the 2026–27 CBSE syllabus, Optics and Electromagnetic Waves carry a combined weightage of 18 marks. Electromagnetic Induction, Alternating Currents, and Magnetic Effects of Current and Magnetism carry 17 marks together, while Electrostatics and Current Electricity carry 16 marks together.
Still, it’s best to prepare all nine units rather than concentrating only on the high-weightage topics.
2. Is NCERT Physics sufficient for CBSE Class 12 Board Exams?
NCERT Physics is the primary textbook prescribed for CBSE Class 12, so understand its concepts, examples, derivations, diagrams and exercises thoroughly. Alongside it, practise official CBSE sample papers, previous-year papers and competency-based questions.
3. How can I improve my numerical problem-solving skills in Physics?
Start by getting the concepts and formulas of each chapter clear. Then practise numericals regularly, convert all quantities into appropriate SI units, show every calculation step, and check the units and dimensions of your final answer.
4. Are derivations important for the Class 12 Physics examination?
Yes. Derivations show how well you understand a concept and how confidently you can use the mathematics. Practise the derivations included in the prescribed syllabus and write them out step by step.
5. How many hours should I study Physics daily?
It depends on your preparation level and your other subjects. A steady daily session of about 1.5–2 hours can work well for concept revision, numerical practice and derivations, and you can increase it as the exam gets closer. If long study stretches leave you tired, the Pomodoro technique can help you break your time into manageable chunks.
6. Where can I download the official CBSE Class 12 Physics sample question paper?
You can download the latest Physics sample question paper and marking scheme from the official CBSE Academic website.
Official resource: https://cbseacademic.nic.in/SQP_CLASSXII_2026-27.html
7. How many questions are there in the CBSE Class 12 Physics theory paper?
The 2026–27 paper has 33 compulsory questions divided into five sections, A to E, for a total of 70 marks.
Final Words
Preparing for CBSE Class 12 Physics gets a lot easier when you have a plan, practise consistently, and focus on understanding the ideas behind the formulas.
The 30 important questions and answers in this article cover the major topics across all nine units of the prescribed syllabus. Use them as a revision resource alongside your NCERT textbooks, classroom notes, previous-year papers and official CBSE sample papers. And if you’d like help planning your revision, our CBSE exam preparation tips are worth a look.
Remember, Physics becomes easier when you understand the principles behind the formulas and practise applying them in different situations. Stay focused on conceptual clarity, accurate numericals, derivations and good time management, and you will see the difference in your performance.
Best wishes to all Class 12 students for their CBSE Physics Board Examinations!
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