JEE Main 2023 previous year paper practice · PYQ-pattern set

JEE Main 2023 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 · Kinematics

   A particle moves along a straight line such that its position x (in meters) at time t (in seconds) is given by x = 4t² - 2t + 5. What is the velocity of the particle at t = 3 s?

   • A. 22 m/s
   • B. 26 m/s
   • C. 18 m/s
   • D. 20 m/s
   Answer and solution

   Answer: A

   Velocity v = dx/dt = 8t - 2. At t = 3 s, v = 8(3) - 2 = 24 - 2 = 22 m/s.

2. Q2 · Properties of Solids and Liquids

   A U-tube contains water and an immiscible liquid of density ρ. The height of the liquid column is 16 cm and that of water is 12 cm. What is the density of the liquid? (Take density of water = 1000 kg/m³)

   • A. 750 kg/m³
   • B. 800 kg/m³
   • C. 850 kg/m³
   • D. 900 kg/m³
   Answer and solution

   Answer: A

   At equilibrium in a U-tube, pressure at the same horizontal level is equal. Therefore ρ₁h₁g = ρ₂h₂g. So ρ × 16 = 1000 × 12, giving ρ = 12000/16 = 750 kg/m³.

3. Q3 · Oscillations and Waves

   A simple pendulum of length L has a time period T on Earth. If the same pendulum is taken to a planet where the acceleration due to gravity is one-fourth that of Earth, the new time period will be

   • A. T/2
   • B. T
   • C. 2T
   • D. 4T
   Answer and solution

   Answer: C

   Time period of a simple pendulum T = 2π√(L/g). When g becomes g/4, the new period T' = 2π√(L/(g/4)) = 2 × 2π√(L/g) = 2T.

4. Q4 · Electromagnetic Waves

   Which of the following electromagnetic waves has the highest frequency?

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

   Answer: C

   In the electromagnetic spectrum, frequency increases in the order: Radio < Microwave < Infrared < Visible < UV < X-rays < Gamma rays. Among the options, X-rays have the highest frequency.

5. Q5 · Dual Nature of Matter and Radiation

   An electron is accelerated through a potential difference of 150 V. What is the de Broglie wavelength associated with it? (Take me = 9.1 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C, h = 6.63 × 10⁻³⁴ J·s)

   • A. 1.0 Å
   • B. 2.5 Å
   • C. 0.5 Å
   • D. 3.2 Å
   Answer and solution

   Answer: A

   Kinetic energy = eV = 150 × 1.6 × 10⁻¹⁹ J. Momentum p = √(2meK). λ = h/p = h/√(2meV) = 6.63×10⁻³⁴/√(2×9.1×10⁻³¹×1.6×10⁻¹⁹×150) ≈ 1.0×10⁻¹⁰ m = 1.0 Å.

6. Q6 · Chemical Bonding and Molecular Structure

   Among the molecules SF₄, XeF₄, BF₃, and NH₃, the total number of species having sp³d hybridization of the central atom is:

   • A. 0
   • B. 1
   • C. 2
   • D. 3
   Answer and solution

   Answer: B

   SF₄ has sp³d hybridization (4 bond pairs + 1 lone pair). XeF₄ has sp³d² hybridization, BF₃ has sp² hybridization, and NH₃ has sp³ hybridization. Only SF₄ has sp³d.

7. Q7 · Equilibrium

   For the gaseous equilibrium PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), 3.0 moles of PCl₅ are introduced into a 2.0 L flask. At equilibrium at 500 K, the concentration of Cl₂ is found to be 0.6 M. What is the value of Kc for this reaction?

   • A. 0.20
   • B. 0.40
   • C. 0.50
   • D. 0.60
   Answer and solution

   Answer: B

   Initial [PCl₅]=1.5 M. Change = 0.6 M (from Cl₂). Equilibrium: PCl₅=0.9, PCl₃=0.6, Cl₂=0.6. Kc=(0.6×0.6)/0.9=0.40.

8. Q8 · p-Block Elements (Groups 13-18)

   White phosphorus reacts with chlorine in limited supply to form PCl₃. The geometry and hybridization of phosphorus in PCl₃ are:

   • A. Trigonal planar, sp²
   • B. Pyramidal, sp³
   • C. Tetrahedral, sp³
   • D. T-shaped, sp³d
   Answer and solution

   Answer: B

   PCl₃ has 3 bond pairs and 1 lone pair on P. Steric number = 4 gives sp³ hybridization. The molecular geometry is pyramidal due to lone pair repulsion.

9. Q9 · Principles Related to Practical Chemistry

   In Kjeldahl's method for estimation of nitrogen, 0.35 g of an organic compound required 30 mL of N/10 H₂SO₄ for complete neutralization of ammonia. The percentage of nitrogen in the compound is approximately:

   • A. 12%
   • B. 24%
   • C. 6%
   • D. 18%
   Answer and solution

   Answer: A

   N/10 H2SO4, 30 mL gives 3 milliequivalents. Nitrogen mass = 3×10⁻³ × 14 = 0.042 g. Percentage = (0.042/0.35)×100 = 12%.

10. Q10 · Matrices and Determinants

    If A = [[2, 3], [1, 2]] and A² - kA + I = O (where I is the 2×2 identity matrix and O is the 2×2 null matrix), then the value of k is:

    • A. 3
    • B. 4
    • C. 5
    • D. 6
    Answer and solution

    Answer: B

    Using the Cayley-Hamilton theorem, A satisfies its own characteristic equation. The characteristic equation is |A - λI| = 0, giving λ² - 4λ + 1 = 0. Hence k equals the trace of A, which is 2 + 2 = 4.