KCET 2025 previous year paper practice · PYQ-pattern set

KCET 2025 PYQ-pattern questions

Showing 10 of the 18 PYQ-pattern questions in this set — freshly worded in that year's pattern, not the original questions. Answers and solutions open under each question.

1. Q1 · Motion in a Straight Line (1st PUC)

   A car moving with uniform acceleration covers 450 m in a 5 s interval, and 700 m in the next 5 s interval. The acceleration of the car is:

   • A. 5 m/s²
   • B. 10 m/s²
   • C. 15 m/s²
   • D. 20 m/s²
   Answer and solution

   Answer: B

   The difference in distance covered in equal time intervals under constant acceleration is Δs = at². Here, 700 - 450 = 250 m and t = 5 s. Using a = Δs/t² = 250/25 = 10 m/s².

2. Q2 · Gravitation (1st PUC)

   The ratio of escape velocity at the surface of a planet to that at Earth is 2:1. If the radius of the planet is equal to that of Earth, the ratio of their masses (M_planet/M_Earth) is:

   • A. 1:1
   • B. 2:1
   • C. 4:1
   • D. 1:4
   Answer and solution

   Answer: C

   Escape velocity v_e = √(2GM/R). Since radii are equal, v_e ∝ √M. Given v_planet/v_Earth = 2, we have √(M_planet/M_Earth) = 2, so M_planet/M_Earth = 4.

3. Q3 · Oscillations (1st PUC)

   A particle executes SHM with amplitude 10 cm and time period 4 s. The time taken by the particle to move from mean position to a point 5 cm away from mean position is:

   • A. 1/3 s
   • B. 1/2 s
   • C. 2/3 s
   • D. 1 s
   Answer and solution

   Answer: A

   Using x = A sin(ωt), we have 5 = 10 sin(2πt/4). This gives sin(πt/2) = 1/2, so πt/2 = π/6, giving t = 1/3 s.

4. Q4 · Electrostatic Potential and Capacitance (2nd PUC)

   Two point charges of +4 μC and -4 μC are placed at points A and B separated by a distance of 8 cm. The electric potential at the midpoint of the line joining A and B is

   • A. Zero
   • B. 9 × 10⁵ V
   • C. 4.5 × 10⁵ V
   • D. 1.8 × 10⁵ V
   Answer and solution

   Answer: A

   At the midpoint, the distances from both charges are equal (4 cm each). Since the charges are equal in magnitude but opposite in sign, their potentials cancel out, resulting in zero net potential.

5. Q5 · Alternating Current (2nd PUC)

   In an AC circuit, a pure inductor of inductance 0.2 H is connected to an AC source of 220 V, 50 Hz. The inductive reactance is (take π = 3.14)

   • A. 31.4 Ω
   • B. 62.8 Ω
   • C. 15.7 Ω
   • D. 125.6 Ω
   Answer and solution

   Answer: B

   Inductive reactance Xₗ = 2πfL = 2 × 3.14 × 50 × 0.2 = 62.8 Ω.

6. Q6 · Wave Optics (2nd PUC)

   In Young's double slit experiment, the slits are separated by 0.5 mm and the screen is placed at a distance of 1.2 m from the slits. If light of wavelength 600 nm is used, what is the fringe width?

   • A. 1.44 mm
   • B. 1.20 mm
   • C. 0.72 mm
   • D. 2.40 mm
   Answer and solution

   Answer: A

   Fringe width β = λD/d = (600×10⁻⁹ × 1.2)/(0.5×10⁻³) = 1.44×10⁻³ m = 1.44 mm.

7. Q7 · Communication Systems (2nd PUC)

   Which of the following frequency ranges corresponds to the microwave portion of the electromagnetic spectrum?

   • A. 1 MHz to 10 MHz
   • B. 1 GHz to 300 GHz
   • C. 10 kHz to 1 MHz
   • D. 300 MHz to 3 GHz
   Answer and solution

   Answer: B

   Microwaves typically occupy the frequency range from about 1 GHz to 300 GHz, used extensively in satellite and radar communication.

8. Q8 · Classification of Elements and Periodicity (1st PUC)

   Which of the following is the correct order of increasing atomic radii?

   • A. F < Cl < Br < I
   • B. I < Br < Cl < F
   • C. Cl < F < Br < I
   • D. Br < Cl < I < F
   Answer and solution

   Answer: A

   Atomic radius increases down a group because a new shell is added. Thus, the correct increasing order is F < Cl < Br < I.

9. Q9 · Redox Reactions (1st PUC)

   In the reaction 2FeCl₃ + SnCl₂ → 2FeCl₂ + SnCl₄, the oxidation state of tin changes from:

   • A. +2 to +4
   • B. +4 to +2
   • C. 0 to +2
   • D. +2 to 0
   Answer and solution

   Answer: A

   Tin in SnCl₂ has oxidation state +2, and in SnCl₄ it is +4. Tin is oxidized from +2 to +4.

10. Q10 · Solid State (2nd PUC)

    A metal crystallizes in a face-centered cubic (fcc) lattice with edge length 408 pm. The radius of the metal atom (in pm) is approximately:

    • A. 102
    • B. 144
    • C. 204
    • D. 288
    Answer and solution

    Answer: B

    In fcc, 4r = a√2, so r = a√2/4 = 408 × 1.414/4 ≈ 144 pm.