JEE Advanced 2021 previous year paper practice · PYQ-pattern set

JEE Advanced 2021 PYQ-pattern questions

Showing 10 of the 24 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 · Modern Physics

   In a photoelectric effect experiment, light of wavelength 400 nm is incident on a metal surface with work function 1.8 eV. The maximum kinetic energy of the ejected photoelectrons is (take h = 6.6 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s, 1 eV = 1.6 × 10⁻¹⁹ J):

   • A. 0.9 eV
   • B. 1.1 eV
   • C. 1.3 eV
   • D. 1.5 eV
   Answer and solution

   Answer: C

   Photon energy = hc/λ = (6.6e-34×3e8)/(400e-9) = 4.95e-19 J = 3.09 eV. KEmax = 3.09 − 1.8 = 1.29 ≈ 1.3 eV.

2. Q2 · Electricity and Magnetism

   A parallel plate capacitor with plate area A and separation d is filled with two dielectrics of equal thickness (d/2 each) having dielectric constants K₁ = 2 and K₂ = 4. If the capacitance with dielectrics is C, the capacitance without dielectrics would be:

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

   Answer: B

   Capacitors in series: 1/C = 1/(K₁ε₀A/(d/2)) + 1/(K₂ε₀A/(d/2)) = (d/2ε₀A)(1/2 + 1/4) = 3d/8ε₀A. Without dielectrics C₀ = ε₀A/d, so C₀ = 3C/8.

3. Q3 · Chemical Thermodynamics

   For a reaction, ΔH = -120 kJ and ΔS = -400 J/K at 298 K. The minimum temperature (in K) above which the reaction becomes non-spontaneous is:

   • A. 300 K
   • B. 298 K
   • C. 350 K
   • D. 400 K
   Answer and solution

   Answer: A

   At equilibrium, ΔG = 0, so T = ΔH/ΔS = -120000/(-400) = 300 K. Above 300 K, TΔS > ΔH (in magnitude), making ΔG positive (non-spontaneous).

4. Q4 · s-Block Elements

   Which of the following alkaline earth metal carbonates is thermally the most stable?

   • A. BeCO₃
   • B. MgCO₃
   • C. CaCO₃
   • D. BaCO₃
   Answer and solution

   Answer: D

   Thermal stability of alkaline earth metal carbonates increases down the group as polarizing power of cation decreases. BaCO₃ is most stable.

5. Q5 · Basic Principles of Organic Chemistry

   Among the following carbocations, which one is most stable due to hyperconjugation and inductive effects combined?

   • A. (CH₃)₃C⁺
   • B. CH₃CH₂CH₂⁺
   • C. C₆H₅CH₂⁺
   • D. CH₂=CH-CH₂⁺
   Answer and solution

   Answer: A

   Tertiary carbocation (CH₃)₃C⁺ is stabilized by nine α-hydrogens providing maximum hyperconjugation and three electron-donating methyl groups via +I effect, making it most stable.

6. Q6 · Coordination Compounds

   The spin-only magnetic moment value (in Bohr magnetons) for the octahedral complex [Fe(CN)₆]³⁻ is approximately:

   • A. 0
   • B. 1.73
   • C. 3.87
   • D. 5.92
   Answer and solution

   Answer: B

   Fe³⁺ has d⁵ configuration. CN⁻ is a strong field ligand causing pairing, giving t₂g⁵ with one unpaired electron. μ = √[n(n+2)] = √[1(3)] = 1.73 BM.

7. Q7 · Aldehydes and Ketones

   Consider the following reaction sequence: Acetone is treated with excess benzaldehyde in the presence of dilute NaOH to give product P. How many α,β-unsaturated carbonyl systems are present in the major product P?

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

   Answer: C

   Acetone undergoes Claisen-Schmidt condensation with excess benzaldehyde at both α-positions, forming dibenzylideneacetone (C₆H₅–CH=CH–CO–CH=CH–C₆H₅), which contains two α,β-unsaturated carbonyl systems.

8. Q8 · Polymers

   Among the following polymers, which one has the strongest intermolecular hydrogen bonding?

   • A. Polyethylene
   • B. Nylon 6,6
   • C. Natural rubber
   • D. Polystyrene
   Answer and solution

   Answer: B

   Nylon 6,6 is a polyamide with extensive intermolecular hydrogen bonding between -NH- and -C=O groups, giving it high tensile strength. Polyethylene, natural rubber, and polystyrene lack such strong hydrogen bonding.

9. Q9 · States of Matter: Gases and Liquids

   At 320 K, the density of a certain gaseous molecule at 3 bar is double that of nitrogen (N₂) at 6 bar at the same temperature. The molar mass of the gaseous molecule (in g/mol) is:

   • A. 28
   • B. 42
   • C. 56
   • D. 112
   Answer and solution

   Answer: D

   From ideal gas law, ρ = PM/(RT). Given ρ₁/ρ₂ = 2 = (P₁M₁)/(P₂M₂). So M₁ = 2×(P₂/P₁)×M₂ = 2×(6/3)×28 = 112 g/mol.

10. Q10 · Chemical Kinetics

    A first-order reaction has a rate constant of 0.0693 min⁻¹. What percentage of the reactant will remain after 20 minutes?

    • A. 25%
    • B. 12.5%
    • C. 6.25%
    • D. 50%
    Answer and solution

    Answer: A

    For first-order: k = 0.0693 min⁻¹ gives half-life t½ = 0.693/0.0693 = 10 min. In 20 minutes = 2 half-lives, so remaining fraction = (1/2)² = 25%.