KCET 2024 previous year paper practice · PYQ-pattern set

KCET 2024 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 · Physical World (1st PUC)

   Which of the following quantities has the same dimensions as force per unit area?

   • A. Energy density
   • B. Linear momentum
   • C. Angular momentum
   • D. Coefficient of viscosity
   Answer and solution

   Answer: A

   Force per unit area is pressure, which has dimensions [ML⁻¹T⁻²]. Energy density (energy/volume) also has dimensions [ML⁻¹T⁻²], making them dimensionally equivalent.

2. Q2 · Work, Energy and Power (1st PUC)

   A spring of force constant 800 N/m is compressed by 10 cm. The potential energy stored in the spring is:

   • A. 2 J
   • B. 4 J
   • C. 6 J
   • D. 8 J
   Answer and solution

   Answer: B

   Potential energy in spring U = ½kx². Here, k = 800 N/m and x = 10 cm = 0.1 m. So U = ½ × 800 × (0.1)² = 400 × 0.01 = 4 J.

3. Q3 · Waves (1st PUC)

   Two waves of equal frequency and amplitude traveling in opposite directions superpose to produce a standing wave. The distance between two consecutive nodes is 20 cm. The wavelength of the component waves is:

   • A. 10 cm
   • B. 20 cm
   • C. 40 cm
   • D. 80 cm
   Answer and solution

   Answer: C

   In a standing wave, the distance between two consecutive nodes is λ/2. Given that this distance is 20 cm, we have λ/2 = 20 cm, therefore λ = 40 cm.

4. Q4 · Electromagnetic Waves (2nd PUC)

   Which of the following electromagnetic waves has the shortest wavelength?

   • A. X-rays
   • B. Ultraviolet rays
   • C. Infrared rays
   • D. Microwaves
   Answer and solution

   Answer: A

   In the electromagnetic spectrum, X-rays have shorter wavelength than UV, IR, and microwaves. Only gamma rays have shorter wavelength than X-rays.

5. Q5 · Dual Nature of Radiation and Matter (2nd PUC)

   The work function of a metal is 2.5 eV. Light of wavelength 400 nm is incident on the metal surface. What is the maximum kinetic energy of the emitted photoelectrons? (Take h = 6.63×10⁻³⁴ Js, c = 3×10⁸ m/s, 1 eV = 1.6×10⁻¹⁹ J)

   • A. 0.6 eV
   • B. 1.2 eV
   • C. 0.8 eV
   • D. 1.5 eV
   Answer and solution

   Answer: A

   Photon energy = hc/λ = 4.97e-19 J = 3.11 eV. Max KE = photon energy - work function = 3.11 - 2.5 = 0.61 eV ≈ 0.6 eV.

6. Q6 · Some Basic Concepts of Chemistry (1st PUC)

   What is the mass of CO₂ produced when 6.0 g of carbon is completely burnt in excess oxygen? (Atomic mass: C=12, O=16)

   • A. 11 g
   • B. 22 g
   • C. 44 g
   • D. 33 g
   Answer and solution

   Answer: B

   Moles of C = 6/12 = 0.5 mol. From C + O₂ → CO₂, 0.5 mol C produces 0.5 mol CO₂. Mass of CO₂ = 0.5×44 = 22 g.

7. Q7 · Chemical Bonding and Molecular Structure (1st PUC)

   The hybridization of carbon in CH₄, C₂H₄ and C₂H₂ are respectively:

   • A. sp³, sp², sp
   • B. sp², sp³, sp
   • C. sp, sp², sp³
   • D. sp³, sp, sp²
   Answer and solution

   Answer: A

   CH₄ has sp³ hybridization (tetrahedral), C₂H₄ has sp² (trigonal planar), and C₂H₂ has sp hybridization (linear).

8. Q8 · Hydrogen (1st PUC)

   Hard water contains dissolved salts of:

   • A. Calcium and magnesium
   • B. Sodium and potassium
   • C. Iron and manganese
   • D. Aluminium and zinc
   Answer and solution

   Answer: A

   Hard water primarily contains dissolved salts of calcium and magnesium, typically as bicarbonates, chlorides, or sulfates.

9. Q9 · p-Block Elements — Groups 13 and 14 (1st PUC)

   A compound of boron reacts with ammonia on heating to give a compound known as 'inorganic graphite'. The boron compound is:

   • A. B₂O₃
   • B. BCl₃
   • C. B₂H₆
   • D. BF₃
   Answer and solution

   Answer: C

   Inorganic graphite is borazine (B₃N₃H₆) or borazon-related; borazine is formed when diborane (B₂H₆) reacts with ammonia on heating. 3 B₂H₆ + 6 NH₃ → 2 B₃N₃H₆ + 12 H₂. This layered borazine structure is called inorganic benzene/graphite analog.

10. Q10 · Electrochemistry (2nd PUC)

    The conductivity of 0.02 M KCl solution is 2.76 × 10⁻³ S cm⁻¹. Its molar conductivity (in S cm² mol⁻¹) is:

    • A. 55.2
    • B. 138
    • C. 276
    • D. 552
    Answer and solution

    Answer: B

    Molar conductivity Λₘ = κ × 1000 / M = (2.76 × 10⁻³ × 1000) / 0.02 = 138 S cm² mol⁻¹.