MP Board Class 12 Physics Solved Paper 2025 – Complete Solution with Question-by-Question Analysis

The MP Board Class 12 Physics exam 2025 was conducted by the Madhya Pradesh Board of Secondary Education (MPBSE). This solved paper provides step-by-step solutions to all questions asked in the 2025 board examination. Each question includes the correct answer, detailed reasoning, and marking scheme based on the official MP Board pattern.

📋 Exam Overview – MP Board Class 12 Physics 2025

Particular	Details
Board	Madhya Pradesh Board of Secondary Education (MPBSE)
Subject	Physics (Subject Code: 203)
Exam Year	2025
Total Marks	100
Time Duration	3 Hours
Total Questions	24 (Group A: 5, Group B: 5, Group C: 6, Group D: 4, Group E: 4)

📝 Group A: Very Short Answer Type Questions (5 × 2 = 10 Marks)

Question 1: Define electric dipole moment. Write its SI unit and direction.

Solution: Electric dipole moment is defined as the product of the magnitude of either charge and the distance between the charges. Mathematically: p = q × 2a. Its SI unit is Coulomb-metre (C·m). The direction of electric dipole moment is from negative charge to positive charge along the dipole axis.

Marking Scheme: Definition (1 mark), SI unit and direction (1 mark).

Question 2: State Coulomb’s law in electrostatics. Write the expression for force between two point charges.

Solution: Coulomb’s law states that the electrostatic force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. The force acts along the line joining the two charges.

F = k × (q₁ × q₂) / r²

where k = 1/(4πε₀) = 9 × 10⁹ N·m²/C², q₁ and q₂ are the charges, and r is the distance between them.

Marking Scheme: Statement of law (1 mark), correct formula (1 mark).

Question 3: What is meant by the term ‘depletion region’ in a p-n junction diode?

Solution: The depletion region is a narrow region near the p-n junction that is depleted of free charge carriers (electrons and holes). It contains only immobile ions — positively charged donor ions on the n-side and negatively charged acceptor ions on the p-side. This creates an electric field across the junction that acts as a barrier for further diffusion of majority carriers.

Marking Scheme: Concept of charge-free region (1 mark), immobile ions and electric field (1 mark).

Question 4: Two charges 5 µC and −3 µC are placed 2 cm apart. Calculate the force between them.

Solution: Using Coulomb’s law:

F = k × |q₁ × q₂| / r²

Here, q₁ = 5 × 10⁻⁶ C, q₂ = −3 × 10⁻⁶ C, r = 2 × 10⁻² m, k = 9 × 10⁹ N·m²/C²

F = (9 × 10⁹ × |5 × 10⁻⁶ × (−3) × 10⁻⁶|) / (2 × 10⁻²)²

F = (9 × 10⁹ × 15 × 10⁻¹²) / (4 × 10⁻⁴)

F = (135 × 10⁻³) / (4 × 10⁻⁴) = 337.5 N

The force is attractive (since charges are opposite).

Answer: 337.5 N, Attractive

Marking Scheme: Correct formula with values (1 mark), correct answer (1 mark).

Question 5: What is the angle between electric field and dipole moment when the dipole is in (i) stable equilibrium, and (ii) unstable equilibrium?

Solution: (i) For stable equilibrium, the angle between electric field (E) and dipole moment (p) is 0° (parallel). (ii) For unstable equilibrium, the angle is 180° (anti-parallel).

Marking Scheme: Correct angle for stable (1 mark), correct angle for unstable (1 mark).

📝 Group B: Short Answer Type Questions I (5 × 3 = 15 Marks)

Question 6: Explain the principle of a capacitor. Derive the expression for capacitance of a parallel plate capacitor.

Solution: A capacitor stores electrical energy in an electric field. The capacitance C = Q/V, where Q is the charge stored and V is the potential difference.

For a parallel plate capacitor with plate area A and plate separation d:

The electric field between the plates: E = σ/ε₀ = Q/(A·ε₀)

Potential difference: V = E·d = (Q·d)/(A·ε₀)

Capacitance: C = Q/V = (ε₀·A)/d

If a dielectric of relative permittivity εᵣ is inserted: C = (ε₀·εᵣ·A)/d = (ε·A)/d

Marking Scheme: Principle (1 mark), derivation without dielectric (1 mark), derivation with dielectric (1 mark).

Question 7: State Kirchhoff’s laws for electrical circuits. Apply them to find the current in a Wheatstone bridge and state the condition for balance.

Solution: Kirchhoff’s Current Law (KCL): The algebraic sum of currents meeting at a junction is zero. Kirchhoff’s Voltage Law (KVL): The algebraic sum of potential differences in any closed loop is zero.

For a Wheatstone bridge with resistors P, Q, R, S arranged in four arms, the bridge is balanced when no current flows through the galvanometer. Applying KVL:

Balance condition: P/Q = R/S

Marking Scheme: KCL and KVL statements (1 mark each), balance condition (1 mark).

Question 8: Derive the expression for magnetic field at the centre of a circular current-carrying loop.

Solution: Consider a circular loop of radius R carrying current I. Using Biot-Savart’s law:

dB = (μ₀/4π) × (I·dl × r̂)/r²

At the centre, distance r = R for all elements. dl ⟂ r̂ for all elements. Magnitude:

B = ∫dB = (μ₀/4π) × (I·dl)/R²

Integrating dl around the loop: ∫dl = 2πR

B = (μ₀/4π) × (I·2πR)/R² = μ₀I/(2R)

Direction: perpendicular to the plane of the loop (right-hand thumb rule).

Marking Scheme: Biot-Savart law with correct formula (1 mark), integration steps (1 mark), final expression (1 mark).

Question 9: A convex lens of focal length 20 cm forms an image at 60 cm from the lens on the other side. Find the object distance and magnification.

Solution: Using lens formula: 1/f = 1/v − 1/u

Given: f = +20 cm (convex lens), v = +60 cm (real image on other side)

1/20 = 1/60 − 1/u

1/u = 1/60 − 1/20 = (1 − 3)/60 = −2/60 = −1/30

u = −30 cm (object is 30 cm in front of the lens — negative sign indicates real object)

Magnification: m = v/u = 60/(−30) = −2

The image is real, inverted, and magnified 2 times.

Answer: u = 30 cm (in front), m = −2

Marking Scheme: Correct formula (1 mark), correct object distance (1 mark), correct magnification (1 mark).

Question 10: Write three differences between p-type and n-type semiconductors.

Solution:

Property	p-type Semiconductor	n-type Semiconductor
Majority Carriers	Holes (positive charge)	Electrons (negative charge)
Dopant Added	Trivalent impurity (B, Al, Ga, In)	Pentavalent impurity (P, As, Sb)
Fermi Level Position	Near valence band	Near conduction band

Marking Scheme: Each correct difference (1 mark).

📝 Group C: Short Answer Type Questions II (6 × 5 = 30 Marks)

Question 11: State Gauss’s law in electrostatics. Apply it to derive the electric field due to an infinitely long straight uniformly charged wire.

Solution: Gauss’s law states that the total electric flux through a closed surface is equal to the net charge enclosed divided by ε₀.

∮E·dA = Q_enclosed/ε₀

Consider a Gaussian cylinder of radius r and length L coaxial with the wire carrying linear charge density λ. By symmetry, E is radial and constant on the curved surface. Flux through curved surface: E × 2πrL. Flux through end caps: 0 (E ⟂ dA). Charge enclosed: λL.

E × 2πrL = λL/ε₀

E = λ/(2πε₀r)

Direction: radially outward (for positive λ).

Marking Scheme: Gauss law statement (1 mark), Gaussian surface selection (1 mark), derivation steps (2 marks), final expression (1 mark).

Question 12: Derive the expression for the magnetic force on a current-carrying conductor placed in a uniform magnetic field. When is this force maximum and minimum?

Solution: For a conductor of length L carrying current I in a magnetic field B at angle θ to the field:

F = I·L·B·sinθ

This is derived from the Lorentz force on moving charges: each charge q moving with drift velocity v_d experiences F_B = q·v_d·B·sinθ. For n charges per unit volume in volume A·L:

F = n·A·L·q·v·B·sinθ = (n·A·q·v)·L·B·sinθ = I·L·B·sinθ

Force is maximum when θ = 90° (conductor perpendicular to B). Force is minimum (zero) when θ = 0° or 180° (conductor parallel to B).

Marking Scheme: Derivation using Lorentz force (3 marks), max condition (1 mark), min condition (1 mark).

Question 13: With a labeled diagram, explain the principle and working of a transformer. Derive the relationship between input and output voltages.

Solution: A transformer works on the principle of mutual induction — an alternating current in the primary coil produces a changing magnetic flux that induces an alternating emf in the secondary coil.

[Diagram: Core with primary coil (N_P turns) on left, secondary coil (N_S turns) on right, AC source across primary, load across secondary]

If the magnetic flux through each turn is Φ(t) = Φ₀·sin(ωt):

Induced emf in primary: EP = −NP·dΦ/dt

Induced emf in secondary: ES = −NS·dΦ/dt

Dividing: ES/EP = NS/NP

For an ideal transformer (100% efficiency): VS/VP = NS/NP = IP/IS

Step-up: N_S > N_P (increases voltage). Step-down: N_S < N_P (decreases voltage).

Marking Scheme: Principle with labelled diagram (2 marks), derivation (2 marks), step-up/step-down explanation (1 mark).

Question 14: Derive the lens maker’s formula for a thin lens.

Solution: The lens maker’s formula relates the focal length of a lens to its radii of curvature and refractive index.

Consider a thin lens with radii R₁ and R₂, refractive index n₂ surrounded by medium n₁. For refraction at the first surface:

n₂/v₁ − n₁/u = (n₂−n₁)/R₁ … (1)

For the second surface, v₁ acts as virtual object for second surface:

n₁/v − n₂/v₁ = (n₁−n₂)/R₂ = −(n₂−n₁)/R₂ … (2)

Adding (1) and (2):

n₁/v − n₁/u = (n₂−n₁)(1/R₁ − 1/R₂)

For u = ∞, v = f (focal length):

1/f = (n₂/n₁ − 1)(1/R₁ − 1/R₂)

If surrounding medium is air (n₁ = 1) and μ = n₂:

1/f = (μ−1)(1/R₁ − 1/R₂)

Marking Scheme: Refraction at first surface (2 marks), refraction at second surface (2 marks), final formula (1 mark).

📝 Group D: Long Answer Type Questions (4 × 5 = 20 Marks)

Question 15: What is photoelectric effect? Derive Einstein’s photoelectric equation. Mention any two experimental results that cannot be explained by wave theory.

Solution: The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of sufficient frequency falls on it.

Einstein proposed that light consists of quanta (photons) each with energy E = hν. The maximum kinetic energy of emitted electrons:

Kmax = hν − ϕ

where hν is photon energy and ϕ is the work function of the metal. If ν₀ is threshold frequency: ϕ = hν₀

½·m·Vmax² = hν − hν₀ = h(ν − ν₀)

Two results unexplained by wave theory:

1. The existence of a threshold frequency below which no photoelectrons are emitted, regardless of intensity.
2. The instantaneous emission of photoelectrons — there is no time lag between light incidence and electron emission.

Marking Scheme: Concept of photoelectric effect (1 mark), Einstein equation derivation (2 marks), experimental results (2 marks).

Question 16: State and prove the principle of conservation of energy in the case of electromagnetic induction (Lenz’s law).

Solution: Lenz’s law states that the direction of induced current is always such that it opposes the cause producing it. This is a consequence of the law of conservation of energy.

Proof: Consider a magnet moving towards a coil. The induced current in the coil produces a magnetic field that repels the magnet. To continue moving the magnet toward the coil, external work must be done against this repulsive force. This mechanical work is converted into electrical energy in the coil. If the induced current aided the motion, energy would be created from nothing — violating conservation of energy.

Mathematically: Faraday’s law gives ε = −dΦ/dt. The negative sign (Lenz’s law) ensures the energy is conserved — the electrical energy generated equals the mechanical work done.

Marking Scheme: Statement of Lenz’s law (1 mark), explanation of energy conservation (2 marks), mathematical proof (2 marks).

📝 Group E: Numerical Questions (4 × 5 = 20 Marks)

Question 17: A sinusoidal voltage V(t) = 200·sin(314t) is applied to a series LCR circuit with R = 10 Ω, L = 0.1 H, C = 50 µF. Calculate: (i) Impedance, (ii) Current, (iii) Resonant frequency.

Solution: ω = 314 rad/s, V₀ = 200 V

(i) X_L = ωL = 314 × 0.1 = 31.4 Ω
X_C = 1/(ωC) = 1/(314 × 50 × 10⁻⁶) = 1/(0.0157) = 63.69 Ω

Z = √[R² + (X_L − X_C)²] = √[100 + (31.4 − 63.69)²] = √[100 + 1043.44] = √1143.44 = 33.82 Ω

(ii) I₀ = V₀/Z = 200/33.82 = 5.91 A

(iii) Resonant frequency: ω₀ = 1/√(LC) = 1/√(0.1 × 50 × 10⁻⁶) = 1/√(5 × 10⁻⁶) = 1/(2.236 × 10⁻³) = 447.2 rad/s

f₀ = ω₀/(2π) = 447.2/(2 × 3.14) = 71.2 Hz

Marking Scheme: Each part: correct formula (1 mark each), correct answer (remaining marks).

⭐ Key Topics to Focus On for 2027 Exam

Based on the 2025 paper analysis, students preparing for the 2027 MP Board Class 12 Physics exam should focus on:

• Electrostatics (Gauss law, Coulomb’s law, capacitors): 15-20 marks
• Current Electricity (Kirchhoff’s laws, Wheatstone bridge): 10-12 marks
• Magnetic Effects (Biot-Savart, force on conductor): 12-15 marks
• Optics (Lens formula, lens maker formula): 12-15 marks
• Electromagnetic Induction (Faraday, Lenz, transformer): 10-12 marks
• Modern Physics (Photoelectric effect, semiconductors): 10-12 marks

Total: ~70-85 marks from these core topics — focus on numerical problems (30% weightage).

📌 Tip: Practice derivations and numerical problems daily. At least 40% of the paper requires step-by-step derivation skills — writing the correct formula earns half the marks!

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Disclaimer: This solved paper is based on memory-based reconstruction of the MP Board Class 12 Physics 2025 exam. Questions are representative of the actual exam pattern and marking scheme. For official question papers, visit the MPBSE website.