KCET 2022 previous year paper practice · PYQ-pattern set

KCET 2022 PYQ-pattern questions

Showing 10 of the 17 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 Plane (1st PUC)

   A projectile is fired at an angle of 30° with the horizontal with velocity 40 m/s. The maximum height attained by the projectile is: (g = 10 m/s²)

   • A. 10 m
   • B. 20 m
   • C. 30 m
   • D. 40 m
   Answer and solution

   Answer: B

   Maximum height H = (u²sin²θ)/(2g). Here, u = 40 m/s, θ = 30°. So H = (40² × sin²30°)/(2 × 10) = (1600 × 0.25)/20 = 20 m.

2. Q2 · Units and Measurement (1st PUC)

   If the unit of length is doubled and the unit of time is halved, the unit of velocity becomes:

   • A. Same as before
   • B. Doubled
   • C. Four times
   • D. Half
   Answer and solution

   Answer: C

   Velocity = length/time. If length unit doubles and time unit is halved, new velocity unit = (2L)/(T/2) = 4(L/T), which is four times the original.

3. Q3 · Thermodynamics (1st PUC)

   An ideal gas undergoes a cyclic process ABCA as shown. The work done by the gas in the complete cycle is 600 J. If heat absorbed along AB is 1200 J and heat rejected along CA is 800 J, the heat exchanged along BC is:

   • A. 200 J absorbed
   • B. 200 J rejected
   • C. 400 J absorbed
   • D. 400 J rejected
   Answer and solution

   Answer: A

   In a complete cycle ΔU=0, so net heat equals net work: Q_AB + Q_BC + Q_CA = W. With 1200 - 800 + Q_BC = 600, Q_BC = 200 J. Positive value means heat is absorbed.

4. Q4 · Magnetism and Matter (2nd PUC)

   A bar magnet of magnetic moment 2.5 A·m² is placed in a uniform magnetic field of 0.4 T making an angle of 30° with the field. The torque acting on the magnet is

   • A. 0.5 N·m
   • B. 1.0 N·m
   • C. 0.25 N·m
   • D. 2.0 N·m
   Answer and solution

   Answer: A

   Torque τ = MB sinθ = 2.5 × 0.4 × sin30° = 2.5 × 0.4 × 0.5 = 0.5 N·m.

5. Q5 · Nuclei (2nd PUC)

   A radioactive sample has a half-life of 4 years. What fraction of the sample will remain undecayed after 12 years?

   • A. 1/4
   • B. 1/8
   • C. 1/16
   • D. 1/2
   Answer and solution

   Answer: B

   Number of half-lives n = 12/4 = 3. Remaining fraction = (1/2)³ = 1/8.

6. Q6 · Wave Optics (2nd PUC)

   Two polaroid sheets are placed such that their pass axes make an angle of 60° with each other. If unpolarized light of intensity I₀ is incident on the first polaroid, what is the intensity of light emerging from the second polaroid?

   • A. I₀/8
   • B. I₀/4
   • C. 3I₀/8
   • D. I₀/2
   Answer and solution

   Answer: A

   After first polaroid: I₁ = I₀/2. After second polaroid using Malus law: I₂ = I₁cos²(60°) = (I₀/2)×(1/4) = I₀/8.

7. Q7 · Thermodynamics (1st PUC)

   The enthalpy change for the reaction C(graphite) + O₂(g) → CO₂(g) is -393.5 kJ mol⁻¹. This value represents:

   • A. Enthalpy of formation of CO₂
   • B. Enthalpy of combustion of carbon
   • C. Both A and B
   • D. Neither A nor B
   Answer and solution

   Answer: C

   The reaction represents both the formation of CO₂ from its elements and the combustion of carbon, so both options A and B are correct.

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

   The first ionization enthalpy of magnesium is higher than that of sodium because:

   • A. Magnesium has a smaller atomic radius
   • B. Magnesium has higher nuclear charge
   • C. Both A and B
   • D. Magnesium has one more electron in the same shell
   Answer and solution

   Answer: C

   Magnesium has higher nuclear charge (Z=12) than sodium (Z=11) and smaller atomic radius, both factors increase ionization enthalpy.

9. Q9 · Hydrocarbons (1st PUC)

   Which of the following alkenes will show geometrical isomerism?

   • A. 2-Methylpropene
   • B. 2-Butene
   • C. Propene
   • D. 1-Butene
   Answer and solution

   Answer: B

   2-Butene (CH₃CH=CHCH₃) has two different groups on each carbon of the double bond, enabling cis-trans isomerism.

10. Q10 · Chemical Kinetics (2nd PUC)

    For a first-order reaction, the time required for 75% completion is 60 minutes. The half-life of the reaction (in minutes) is:

    • A. 20
    • B. 30
    • C. 40
    • D. 45
    Answer and solution

    Answer: B

    For first-order: t₇₅% = 2 × t₁/₂. Given t₇₅% = 60 min, so t₁/₂ = 60/2 = 30 minutes.