Karnataka 2nd PUC Physics Model Paper 2027 — Full Paper With Solutions
Complete Karnataka 2nd PUC Physics model paper 2027 with all sections, detailed solutions and chapter-wise weightage. Based on official KSEAB blueprint for board exam preparation.
This model paper follows the official KSEAB 2nd PUC Physics pattern for 2027. Theory: 70 marks in 3 hours 15 minutes. Practical: 30 marks separately.
Paper Structure
| Section | Type | Questions | Marks |
|---|---|---|---|
| Part A | MCQ + Fill in Blanks | 10 | 10 |
| Part B | 2-mark questions (6 of 9) | 9 | 12 |
| Part C | 3-mark questions (6 of 9) | 9 | 18 |
| Part D | 5-mark numericals (3 of 5) | 5 | 15 |
| Part E | 5-mark + 3-mark questions (1 of 2) | 2 | 8+5=... wait |
PART A — Objective (1 Mark Each)
Q1. The SI unit of electric field intensity is:
(a) N (b) N/C (c) C/m (d) V·m
Answer: (b) N/C
Q2. The process of converting AC to DC is called:
(a) Amplification (b) Modulation (c) Rectification (d) Oscillation
Answer: (c) Rectification
Q3. A wire of resistance R is stretched to double its length. Its new resistance is:
(a) R/2 (b) R (c) 2R (d) 4R
Answer: (d) 4R
(When length doubles, area halves. R = ρL/A → R' = ρ(2L)/(A/2) = 4R)
Q4. The photoelectric effect proves the ________ nature of light.
Answer: Particle (quantum)
Q5. The critical angle for total internal reflection depends on:
(a) Wavelength of light (b) Refractive index of medium (c) Intensity (d) Both a and b
Answer: (d) Both a and b
PART B — 2-Mark Questions (Attempt 6 of 9)
Q6. State Coulomb's Law. Write its mathematical form.
Answer: The electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
F = kq₁q₂/r²
where k = 9×10⁹ N·m²/C², q₁, q₂ are charges, r is distance between them.
Q7. What is meant by threshold frequency in photoelectric effect?
Answer: The minimum frequency of incident light below which no photoelectric emission occurs, regardless of the intensity of light, is called the threshold frequency (ν₀).
Below ν₀ — no electrons emitted even with very bright light.
Above ν₀ — electrons are emitted even with dim light.
Q8. State Lenz's Law.
Answer: The direction of the induced current is such that it opposes the change in magnetic flux that causes it.
Lenz's Law is a consequence of the law of conservation of energy.
If it were not so, energy would be created from nothing — which is impossible.
Q9. Define electric potential. Write its SI unit.
Answer: Electric potential at a point is defined as the work done per unit positive charge in bringing a test charge from infinity to that point.
V = W/q
SI unit: Volt (V) = Joule/Coulomb
Potential is a scalar quantity.
Q10. What is a p-n junction diode? Mention its two types of biasing.
Answer: A p-n junction diode is formed by joining p-type and n-type semiconductors together. A depletion layer forms at the junction.
Two types of biasing:
- Forward bias — p-type connected to +ve terminal; depletion layer narrows; current flows
- Reverse bias — p-type connected to −ve terminal; depletion layer widens; no current flows
PART C — 3-Mark Questions (Attempt 6 of 9)
Q11. Derive the expression for the equivalent resistance of resistors connected in parallel.
Answer:
Let R₁, R₂, R₃ be resistors connected in parallel with potential difference V across each.
Current through each: I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃
Total current: I = I₁ + I₂ + I₃
V/Req = V/R₁ + V/R₂ + V/R₃
Dividing by V:
1/Req = 1/R₁ + 1/R₂ + 1/R₃
The equivalent resistance in parallel is always less than the smallest individual resistance.
Q12. Explain the working principle of a transformer. Derive the turns ratio equation.
Answer:
Principle: Mutual induction — changing current in primary coil induces EMF in secondary coil through changing magnetic flux.
Working:
- AC supply to primary coil (Np turns)
- Alternating magnetic flux created in iron core
- Flux links with secondary coil (Ns turns)
- EMF induced in secondary: Es = −Ns(dΦ/dt)
- Primary EMF: Ep = −Np(dΦ/dt)
Turns Ratio: Ep/Es = Np/Ns
Vs/Vp = Ns/Np
- Step-up transformer: Ns > Np → Vs > Vp
- Step-down transformer: Ns < Np → Vs < Vp
Q13. State and explain Huygen's Principle of wave propagation.
Answer:
Huygens' Principle:
- Every point on a given wavefront acts as a source of secondary wavelets which spread out in all directions with the speed of the wave
- The new wavefront at any later time is the forward envelope (tangent) of all the secondary wavelets
Explanation:
If AB is a plane wavefront at time t=0, each point on AB sends out secondary wavelets of radius ct (where c = speed of light).
The tangent drawn to all these spheres gives the new wavefront A'B' at time t.
This principle explains reflection, refraction, and diffraction of waves.
PART D — 5-Mark Numericals (Attempt 3 of 5)
Q14. A parallel plate capacitor has plates of area 6×10⁻³ m² separated by 3 mm of air. Find: (i) Capacitance (ii) Charge if V = 100V (iii) Energy stored.
Solution:
ε₀ = 8.85 × 10⁻¹² F/m
(i) C = ε₀A/d = (8.85×10⁻¹²×6×10⁻³) / (3×10⁻³)
C = 17.7×10⁻¹² F = 17.7 pF
(ii) Q = CV = 17.7×10⁻¹² × 100 = 1.77×10⁻⁹ C = 1.77 nC
(iii) E = ½CV² = ½×17.7×10⁻¹²×(100)²
E = ½×17.7×10⁻¹²×10⁴ = 8.85×10⁻⁸ J
Q15. Light of wavelength 5000 Å falls on a metal with work function 1.9 eV. Find: (i) Energy of photon (ii) Maximum KE of emitted electrons (iii) Stopping potential.
Solution:
h = 6.63×10⁻³⁴ J·s; c = 3×10⁸ m/s; λ = 5×10⁻⁷ m; 1eV = 1.6×10⁻¹⁹ J
(i) E = hc/λ = (6.63×10⁻³⁴×3×10⁸)/(5×10⁻⁷)
E = 3.978×10⁻¹⁹ J = 2.49 eV
(ii) KEmax = E − W₀ = 2.49 − 1.9 = 0.59 eV = 9.44×10⁻²⁰ J
(iii) eV₀ = KEmax → V₀ = 0.59/1 = 0.59 V
Practice Problems (Self Test)
- Three resistors 2Ω, 3Ω, 6Ω are connected in parallel. Find equivalent resistance and total current when 12V is applied.
- A transformer has 500 turns in primary and 2500 turns in secondary. If primary voltage is 220V, find secondary voltage.
- An electron moves with velocity 10⁶ m/s in a magnetic field of 0.5T perpendicular to it. Find the radius of circular path.
Chapter-Wise Weightage
| Chapter | Marks |
|---|---|
| Electrostatics | 9 |
| Current Electricity | 8 |
| Magnetism | 8 |
| Electromagnetic Induction and AC | 8 |
| Optics | 10 |
| Modern Physics | 8 |
| Semiconductor Devices | 7 |
| Communication Systems | 4 |
| Electromagnetic Waves | 3 |
Tip: Karnataka 2nd PUC awards 5 marks for each numerical in Part D. Always write the formula first, substitute values with units, and box your final answer — this ensures full marks even if arithmetic has a small error.
Recommended Resource
Karnataka 2nd PUC Physics Solved Papers — Amazon India
Karnataka PUC Physics solved papers with step-by-step solutions.
Platform: Amazon India · Affiliate link — we earn a small commission at no extra cost to you.