Bihar Board 12th Physics Model Paper 2027 — Full Paper With Solutions

This model paper follows the official BSEB pattern for Class 12th Physics 2027. Theory: 70 marks in 3 hours. Practical: 30 marks.

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Paper Structure

Section	Type	Marks
Section A	MCQ — 35 questions (attempt all)	35
Section B	Short Answer — 15 questions (attempt 10, 3 marks each)	30
Section C	Long Answer — 5 questions (attempt 1, 5 marks)	5

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SECTION A — MCQ (1 Mark Each — Attempt All 35)

Q1. Electric field lines start from ________ charges and end on ________ charges.

(a) negative, positive (b) positive, negative (c) positive, positive (d) negative, negative

Answer: (b) positive, negative

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Q2. The capacitance of a parallel plate capacitor increases when:

(a) Distance between plates increases (b) Area of plates decreases
(c) A dielectric is inserted between plates (d) Voltage across plates increases

Answer: (c) A dielectric is inserted between plates
(C = Kε₀A/d — inserting dielectric increases K, hence C increases)

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Q3. Resistance of a conductor depends on:

(a) Length only (b) Area only (c) Material only (d) All of the above

Answer: (d) All of the above
(R = ρL/A — depends on resistivity ρ, length L, and area A)

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Q4. The phenomenon of electromagnetic induction was discovered by:

(a) Maxwell (b) Oersted (c) Faraday (d) Ampere

Answer: (c) Faraday
(Michael Faraday, 1831)

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Q5. In a transformer, if the number of turns in the secondary coil is greater than in the primary coil, it is called:

(a) Step-down transformer (b) Step-up transformer (c) Isolation transformer (d) Auto transformer

Answer: (b) Step-up transformer

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Q6. The wavelength range of visible light is approximately:

(a) 100–400 nm (b) 400–700 nm (c) 700–1000 nm (d) 1–100 nm

Answer: (b) 400–700 nm
(Violet ~400 nm to Red ~700 nm)

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Q7. Total internal reflection occurs when light passes from:

(a) Rarer to denser medium (b) Denser to rarer medium, angle > critical angle
(c) Any medium at 90° (d) Vacuum to any medium

Answer: (b) Denser to rarer medium, angle > critical angle

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Q8. The de Broglie wavelength of a particle is given by:

(a) λ = hm/v (b) λ = h/mv (c) λ = mv/h (d) λ = hv/m

Answer: (b) λ = h/mv

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Q9. In a nuclear fission reaction, the total mass of products is:

(a) Equal to mass of reactants (b) Greater than mass of reactants
(c) Less than mass of reactants (d) Zero

Answer: (c) Less than mass of reactants
(Mass defect is converted to energy via E = mc²)

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Q10. Which of the following is a semiconductor?

(a) Copper (b) Silicon (c) Silver (d) Mica

Answer: (b) Silicon

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SECTION B — Short Answer (3 Marks Each — Attempt 10 of 15)

Q11. State and explain Gauss's Law in electrostatics.

Answer: Gauss's Law: The total electric flux through any closed surface equals 1/ε₀ times the total charge enclosed.

ΦE = ∮E·dA = q_enc/ε₀

Explanation:

• q_enc = total charge inside the Gaussian surface
• ε₀ = 8.85 × 10⁻¹² C²/N·m² (permittivity of free space)
• The shape of the Gaussian surface does not matter — only the charge enclosed matters
• Useful for calculating electric fields for symmetric charge distributions (sphere, cylinder, plane)

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Q12. Derive the mirror formula: 1/f = 1/v + 1/u

Answer:
Consider a concave mirror with centre of curvature C, focus F, and pole P.
Let object distance = u (negative), image distance = v (negative for real image), focal length = f (negative for concave).

Using sign convention and geometry of similar triangles formed by object, image and mirror:

From △ABC and △A'B'C:
AB/A'B' = BC/B'C ...(i)

From △ABF and △MPF:
AB/MP = BF/FP
Since MP = A'B': AB/A'B' = BF/FP ...(ii)

From (i) and (ii): BC/B'C = BF/FP

Substituting: BC = u−R, B'C = v−R, BF = u−f, FP = f
After substituting R = 2f and simplifying:

1/v + 1/u = 1/f ∎

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Q13. What is photoelectric effect? State Einstein's photoelectric equation.

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

Key observations:

• Below threshold frequency (ν₀), no emission regardless of intensity
• Above ν₀, emission starts instantly
• Kinetic energy of electrons depends on frequency, not intensity
• Intensity increases number of electrons, not their energy

Einstein's Photoelectric Equation:
KEmax = hν − W₀ = hν − hν₀
where:

• h = Planck's constant = 6.63 × 10⁻³⁴ J·s
• ν = frequency of incident light
• W₀ = hν₀ = work function of metal

Stopping potential: eV₀ = KEmax

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Q14. Explain n-type and p-type semiconductors with diagrams.

Answer:

n-type Semiconductor:

• Pure Si or Ge doped with pentavalent impurity (P, As, Sb)
• Impurity atom has 5 valence electrons; 4 form bonds with Si; 1 extra electron is free
• Majority carriers: electrons; Minority carriers: holes
• Net charge is neutral

p-type Semiconductor:

• Pure Si or Ge doped with trivalent impurity (B, Al, Ga, In)
• Impurity atom has 3 valence electrons; creates a hole (missing electron)
• Majority carriers: holes; Minority carriers: electrons
• Net charge is neutral

(Draw Si lattice with P impurity for n-type and B impurity for p-type)

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Q15. Write three applications of total internal reflection.

Answer:

1. Optical Fibres: Light undergoes repeated total internal reflection inside the fibre and travels along it without loss. Used in telecommunications (internet), medical endoscopy, and decorative lighting.

2. Periscope and Binoculars: Totally reflecting prisms are used instead of mirrors. A 45° prism reflects light at 90° by TIR. More efficient than silvered mirrors.

3. Brilliance of Diamond: Diamond is cut so that light entering it undergoes multiple total internal reflections before finally emerging. Critical angle of diamond is only 24.4° — light is trapped and reflected repeatedly, creating the brilliant sparkle.

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SECTION C — Long Answer (5 Marks — Attempt 1)

Q16. Derive an expression for the electric field due to an infinite line charge. Draw the Gaussian surface used.

Answer:
Consider an infinitely long straight wire with linear charge density λ (charge per unit length).

Step 1: Choose Gaussian Surface
By symmetry, E is radial and perpendicular to the wire. Choose a cylindrical Gaussian surface of radius r and length L coaxial with the wire.

Step 2: Calculate Flux

• Flux through curved surface: ΦE = E × 2πrL (E is uniform and perpendicular to area)
• Flux through flat ends = 0 (E is parallel to end faces)
• Total flux = E × 2πrL

Step 3: Apply Gauss's Law
E × 2πrL = q_enc/ε₀ = λL/ε₀

Step 4: Solve for E
E = λ/(2πε₀r)

Direction: Radially outward for positive λ, inward for negative λ.

The electric field decreases as 1/r (not 1/r² as for a point charge).

(Draw cylinder around wire showing E arrows pointing radially outward)

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Practice Problems (Self Test)

1. Two charges +4μC and −4μC are placed 20 cm apart. Find the electric field at the midpoint.
2. A coil of 100 turns and area 0.01 m² is rotated in a magnetic field of 0.5 T at 50 Hz. Find peak EMF.
3. Find the stopping potential for photoelectric effect if frequency of light = 6×10¹⁴ Hz and work function = 2 eV.

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Chapter-Wise Weightage

Chapter	Marks
Electrostatics	10
Current Electricity	8
Magnetic Effects	8
Electromagnetic Induction	8
Optics	12
Modern Physics	10
Semiconductor	7
Communication	4
EM Waves	3

Tip: BSEB 12th Physics Section A (MCQ) is 35 marks — exactly half the paper. Revise all formulas and definitions as one-line facts. Each MCQ tests one concept directly. Covering high-weightage chapters (Optics 12 marks, Electrostatics 10 marks, Modern Physics 10 marks) in Section A alone can secure 25+ marks.