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

JEE Advanced 2023 PYQ-pattern questions

Showing 10 of the 26 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 · General Topics

   The time required for 75% completion of a first-order reaction is 120 minutes. The time required for 93.75% completion of the same reaction (in minutes) is:

   • A. 180
   • B. 210
   • C. 240
   • D. 270
   Answer and solution

   Answer: C

   For first-order reaction: t₇₅% = 2t₁/₂ = 120 min, so t₁/₂ = 60 min. For 93.75% completion (i.e., 6.25% remaining = 1/16), we need 4 half-lives. Therefore t = 4 × 60 = 240 min.

2. Q2 · Solid State

   In a face-centered cubic (FCC) unit cell, the number of atoms per unit cell is:

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

   Answer: C

   In FCC, atoms are at corners (8 × 1/8) and face centers (6 × 1/2). Total = 1 + 3 = 4 atoms per unit cell.

3. Q3 · Surface Chemistry

   Which of the following statements is correct regarding physical adsorption?

   • A. It involves formation of covalent bonds between adsorbate and adsorbent
   • B. It is highly specific and occurs at a fixed temperature only
   • C. The enthalpy change is usually in the range of 20–40 kJ mol⁻¹
   • D. It requires high activation energy to proceed
   Answer and solution

   Answer: C

   Physical adsorption is due to van der Waals forces and typically has low enthalpy of adsorption (20–40 kJ/mol), no specificity, and low activation energy.

4. Q4 · d-Block Elements

   The spin-only magnetic moment (in BM) of the complex ion [Fe(CN)₆]⁴⁻ is closest to:

   • A. 0
   • B. 2.83
   • C. 4.90
   • D. 5.92
   Answer and solution

   Answer: A

   Fe in [Fe(CN)₆]⁴⁻ is in +2 state (d⁶). CN⁻ is a strong field ligand causing pairing. All electrons paired, so n = 0 and μ = 0 BM.

5. Q5 · Principles of Qualitative Analysis

   A salt solution gives a white precipitate with dilute HCl which is soluble in hot water. When H₂S is passed through the acidified solution, a black precipitate is obtained. The cation present is:

   • A. Ag⁺
   • B. Pb²⁺
   • C. Hg²⁺
   • D. Cu²⁺
   Answer and solution

   Answer: B

   Pb²⁺ gives white PbCl₂ precipitate with dilute HCl that dissolves in hot water. When H₂S is passed, black PbS precipitate forms, confirming Pb²⁺.

6. Q6 · Alkenes and Alkynes

   When propyne is treated with sodamide (NaNH₂) followed by ethyl bromide, the major product formed is:

   • A. Pent-1-yne
   • B. Pent-2-yne
   • C. Hex-2-yne
   • D. But-1-yne
   Answer and solution

   Answer: B

   Propyne reacts with NaNH₂ to form sodium propynide (CH₃C≡C⁻Na⁺), which undergoes nucleophilic substitution with ethyl bromide to give CH₃C≡C-CH₂CH₃ (pent-2-yne).

7. Q7 · Alcohols

   A primary alcohol R–CH₂–OH is oxidized using pyridinium chlorochromate (PCC) in dichloromethane. The product formed is further treated with LiAlH₄ followed by H₃O⁺. The final product obtained is:

   • A. R–COOH
   • B. R–CHO
   • C. R–CH₂–OH
   • D. R–CH₃
   Answer and solution

   Answer: C

   PCC oxidizes primary alcohol to aldehyde (R–CHO). LiAlH₄ is a strong reducing agent that reduces aldehyde back to primary alcohol (R–CH₂–OH).

8. Q8 · Chemistry in Everyday Life

   Which of the following statements are correct regarding antacids? (A) They neutralize excess acid in the stomach (B) Ranitidine is an H₂ receptor antagonist (C) They increase the pH of gastric juice (D) Aluminium hydroxide is an example

   • A. A, B and C only
   • B. A, C and D only
   • C. B and D only
   • D. A, B, C and D
   Answer and solution

   Answer: D

   All statements are correct. Antacids neutralize excess acid (A), ranitidine acts on H₂ receptors (B), pH of gastric juice increases (C), and aluminium hydroxide is a common antacid (D).

9. Q9 · Differential Calculus

   Let f : ℝ → ℝ be a differentiable function such that f(x + y) = f(x)f(y) for all x, y ∈ ℝ and f(0) = 1, f'(0) = 3. Then f(2) equals

   • A. e³
   • B. e⁴
   • C. e⁶
   • D. e⁹
   Answer and solution

   Answer: C

   From the functional equation f(x + y) = f(x)f(y) and f(0) = 1, we have f(x) = eᵏˣ for some constant k. Differentiating, f'(x) = keᵏˣ. Since f'(0) = 3, we get k = 3. Therefore f(x) = e³ˣ, and f(2) = e⁶.

10. Q10 · Analytical Geometry — Two Dimensions

    The locus of the midpoint of the chord of the circle x² + y² = 16 which subtends a right angle at the center is

    • A. x² + y² = 4
    • B. x² + y² = 8
    • C. x² + y² = 12
    • D. x² + y² = 16
    Answer and solution

    Answer: B

    For a chord subtending angle 90° at center of radius r = 4, if (h, k) is midpoint, then by geometry OM² = r²/2 where M is midpoint. Thus h² + k² = 16/2 = 8. The locus is x² + y² = 8.