MP Board Class 12 Physics 2023 Previous Year Paper Solved — Question-by-Question Analysis for 2027 Board Exam

📋 Table of Contents

1. Paper Overview
2. Section A — Multiple Choice Questions (1 mark each)
3. Section B — Very Short Answer Questions (2 marks each)
4. Section C — Short Answer Questions (3 marks each)
5. Section D — Long Answer Questions (5 marks each)
6. Expert Preparation Tips for 2027
7. Frequently Asked Questions

💡 Key Insights from the 2023 Paper

• ✨ Electrostatics & Current Electricity — 18 marks (highest weightage)
• ✨ Optics — 15 marks
• ✨ Magnetism & Electromagnetic Induction — 14 marks
• ✨ Modern Physics — 12 marks
• ✨ Numerical questions accounted for 28% of total marks

Q1. Which of the following has the highest dielectric strength?

Answer: (c) Vacuum

Explanation: Dielectric strength is the maximum electric field a material can withstand without breakdown. Vacuum has the highest dielectric strength (~10⁸ V/m) because there are no free charge carriers to initiate ionization. Air has lower dielectric strength (~3 × 10⁶ V/m) due to free electrons and ions.

Q2. The SI unit of magnetic flux is:

• (a) Tesla
• (b) Weber ✓
• (c) Gauss
• (d) Henry

Answer: (b) Weber

Explanation: Magnetic flux Φ = B·A, where B is magnetic field (Tesla) and A is area (m²). So the SI unit is T·m² = Weber (Wb). 1 Wb = 1 T·m². Note: Gauss is the CGS unit of magnetic field (1 T = 10⁴ G). Henry is the SI unit of inductance.

Q3. Which electromagnetic wave is used in remote sensing and radar communication?

Answer: Microwaves

Explanation: Microwaves (wavelength ~1 mm to 30 cm) are ideal for radar and remote sensing because they can penetrate clouds, fog, and rain, and reflect well from metallic objects. Their short wavelength allows high-resolution detection. Applications include weather radar, satellite communication, and microwave ovens.

Q4. In a p-n junction diode, the reverse saturation current depends on:

Answer: Temperature

Explanation: The reverse saturation current I₀ in a p-n junction is given by I₀ = A·T³·exp(−E_g/kT), where T is temperature and E_g is the band gap energy. It approximately doubles for every 10°C rise in temperature due to increased generation of electron-hole pairs in the depletion region.

Q5. Energy equivalent of 1 atomic mass unit (1 u) is:

Answer: 931.5 MeV

Explanation: Using Einstein’s mass-energy equivalence E = mc²: 1 u = 1.6605 × 10⁻²⁷ kg, c = 3 × 10⁸ m/s. E = 1.6605 × 10⁻²⁷ × (3 × 10⁸)² = 1.492 × 10⁻¹⁰ J. Converting to eV: E = 1.492 × 10⁻¹⁰ / 1.602 × 10⁻¹⁹ = 931.5 × 10⁶ eV = 931.5 MeV. This is a fundamental constant used in nuclear physics calculations.

Q6. State Gauss’s law in electrostatics.

Answer: Gauss’s law states that the total electric flux through a closed surface is equal to 1/ε₀ times the net charge enclosed within the surface: ∮ E·dS = Q_enc/ε₀

Q7. Define the term ‘drift velocity’ of electrons in a conductor.

Answer: Drift velocity is the average velocity attained by free electrons in a conductor under the influence of an applied electric field. Its magnitude is typically 10⁻⁴ m/s and is given by v_d = (eE/m)τ, where τ is the average relaxation time between collisions.

Q8. What is the principle of a transformer?

Answer: A transformer works on the principle of mutual induction — when an alternating current flows through the primary coil, it produces a changing magnetic flux through the secondary coil, which induces an EMF in the secondary coil.

Q9. What is meant by ‘polarization of light’?

Answer: Polarization is the process of restricting the vibrations of light waves to a single plane. In unpolarized light, electric field vectors vibrate in all directions perpendicular to propagation. After polarization, the wave vibrates in only one plane (plane-polarized light).

Q10. Define ‘half-life’ of a radioactive substance.

Answer: Half-life (T_1/2) is the time required for a radioactive substance to reduce to half of its initial number of atoms. It is related to the decay constant λ by T_1/2 = ln(2)/λ = 0.693/λ. Half-life is a characteristic property of each radioactive isotope.

Q11. Derive an expression for the electric field due to an electric dipole at a point on its axial line.

Answer: For an electric dipole with charges +q and −q separated by distance 2a, the electric field at a point P on the axial line at distance r from the centre is:

E = (1/4πε₀) × 2p/r³ (for r ≫ a)

where p = q × 2a is the dipole moment vector. The field direction is from negative to positive charge along the dipole axis.

Step-by-step derivation:

1. Field due to +q at P: E₊ = q/(4πε₀(r−a)²) (away from +q)
2. Field due to −q at P: E₋ = q/(4πε₀(r+a)²) (towards −q)
3. Net field E = E₊ − E₋
4. E = q/(4πε₀)[1/(r−a)² − 1/(r+a)²]
5. For r ≫ a: E = (1/4πε₀) × 2q·2a/r³ = (1/4πε₀) × 2p/r³

Q12. State and explain Kirchhoff’s laws for electrical circuits.

Answer: Kirchhoff’s laws consist of two rules:

Junction Rule (KCL): The algebraic sum of currents meeting at a junction is zero. ΣI = 0. This is based on conservation of charge — charge cannot accumulate at a junction.

Loop Rule (KVL): The algebraic sum of potential differences around any closed loop is zero. ΣΔV = 0. This follows conservation of energy — the work done by the electric field in a closed path is zero.

Q13. Explain the principle, construction, and working of a cyclotron.

Answer: A cyclotron accelerates charged particles (protons, deuterons) to high energies using a combination of a magnetic field and an alternating electric field.

Principle: A charged particle moving perpendicular to a uniform magnetic field experiences a centripetal force, making it move in a circular path. The frequency of revolution is independent of speed: f = qB/(2πm).

Construction: Two hollow D-shaped electrodes (dees) placed in a uniform magnetic field perpendicular to their plane. A high-frequency AC voltage is applied between the dees.

Working: A particle from the ion source is accelerated across the gap between dees. Inside each dee, it moves in a semicircular path (no electric field). Every time it crosses the gap, the polarity reverses and accelerates it further. The radius of each half-circle increases, and the particle spirals outward until it exits with maximum energy: Eₘₐₓ = (q²B²R²)/(2m).

Q14. Derive the lens maker’s formula for a thin lens.

Answer: The lens maker’s formula relates the focal length f of a lens to its radii of curvature R₁, R₂ and refractive index n:

1/f = (n − 1)(1/R₁ − 1/R₂)

Using sign convention: R₁ is positive for convex surfaces facing the incident light, R₂ is positive for convex surfaces facing the transmitted light. This formula is derived by applying the refraction formula for spherical surfaces to both surfaces of the lens and combining them.

Q15. Explain the photoelectric effect and state Einstein’s photoelectric equation.

Answer: The photoelectric effect is the emission of electrons from a metal surface when light of suitable frequency falls on it.

Einstein’s photoelectric equation: hν = φ + K_max = hν₀ + ½mv²_max

where hν is the energy of incident photon, φ = hν₀ is the work function (minimum energy to eject electron), and K_max is the maximum kinetic energy of emitted electrons. This equation won Einstein the Nobel Prize in 1921 and established the particle nature of light.

Q16. Explain the construction and working of a moving coil galvanometer. How can it be converted into (a) an ammeter and (b) a voltmeter?

Answer: A moving coil galvanometer is a device used to detect and measure small currents.

Construction: A rectangular coil of many turns wound on a soft iron core, suspended between the poles of a permanent magnet using a phosphor-bronze strip. A pointer attached to the coil moves over a scale.

Working: When current passes through the coil, a torque τ = NIAB acts on it (where N = number of turns, I = current, A = area, B = magnetic field). The restoring torque due to the suspension spring is τ = kθ. At equilibrium, θ = (NAB/k)I, so deflection is proportional to current.

Conversion to Ammeter: Connect a small resistance S (shunt) in parallel with the galvanometer. For current I, shunt S = I_gG/(I − I_g) where I_g is full-scale deflection current and G is galvanometer resistance.

Conversion to Voltmeter: Connect a high resistance R in series with the galvanometer. R = V/I_g − G, where V is the desired voltage range.

Q17. Describe Young’s double-slit experiment. Derive the conditions for constructive and destructive interference.

Answer: Young’s double-slit experiment (YDSE) demonstrated the wave nature of light through interference.

Setup: A monochromatic light source illuminates a single slit. The light then falls on two closely spaced slits S₁ and S₂. The waves from S₁ and S₂ interfere on a screen placed at distance D.

Constructive interference (bright fringe): Path difference = nλ (n = 0, 1, 2, …)
Position: y_n = nλD/d, where d = slit separation.

Destructive interference (dark fringe): Path difference = (2n+1)λ/2
Position: y_n = (2n+1)λD/2d

Fringe width: β = λD/d. This formula is used to determine the wavelength of light experimentally.

• ⭐ Master derivations — Write step-by-step derivations for dipole field, cyclotron, lens maker’s formula, and galvanometer. These are high-weightage topics appearing every year.
• ⭐ Practice numericals daily — Solve at least 5 numerical problems from Electrostatics, Current Electricity, and Optics every day. Focus on formula application and unit conversion.
• ⭐ Use NCERT as primary reference — 90% of MP Board Physics questions come directly from NCERT textbooks. Read each chapter thoroughly and solve all in-text and exercise problems.
• ⭐ Draw diagrams with labels — Well-labeled diagrams (circuit diagrams, ray diagrams, energy band diagrams) can fetch you full marks even if the written explanation is brief.
• ⭐ Solve 5+ previous year papers — Time yourself for each paper (3 hours). This builds speed and confidence. Analyze your weak areas after each test.
• ⭐ Focus on Modern Physics — Topics like photoelectric effect, atomic spectra, nuclear physics are relatively easy and carry 12+ marks. Mastering these can significantly boost your score.

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