JEE Advanced 2024 previous year paper practice · PYQ-pattern set

JEE Advanced 2024 PYQ-pattern questions

Showing 10 of the 28 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 · Electromagnetic Waves

   An electromagnetic wave is propagating in vacuum along the z-direction. The electric field is oscillating with amplitude E₀ = 30 V/m along the x-direction. The amplitude of the magnetic field (in Tesla) is approximately:

   • A. 10⁻⁸
   • B. 10⁻⁷
   • C. 10⁻⁶
   • D. 10⁻⁵
   Answer and solution

   Answer: B

   For EM waves in vacuum, E₀/B₀ = c. Therefore B₀ = E₀/c = 30/(3×10⁸) = 10⁻⁷ T.

2. Q2 · Atomic Structure

   For a hydrogen-like ion (atomic number Z), the wavelength of the radiation emitted when an electron jumps from n = 4 to n = 2 is 121.5 nm. What is the value of Z?

   • A. 1
   • B. 2
   • C. 3
   • D. 4
   Answer and solution

   Answer: B

   Using the Rydberg formula 1/λ = RZ²[1/n₁² - 1/n₂²]. For n = 4→2, 1/λ = RZ²[1/4 - 1/16] = RZ²(3/16). For H (Z=1), n=3→2 gives Hα at 656 nm. For this transition 4→2 and λ = 121.5 nm (close to Lyman α), solving gives Z = 2 (He⁺).

3. Q3 · Chemical Kinetics

   A first-order reaction has a rate constant of 0.0693 min⁻¹. The time required for 75% of the reactant to decompose is:

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

   Answer: B

   For first order, t = (2.303/k) log([A₀]/[A]). For 75% decomposition, [A] = 0.25[A₀]. t = (2.303/0.0693) log(4) = 33.3 × 0.602 ≈ 20 min.

4. Q4 · p-Block Elements

   Among the following oxoacids of phosphorus, how many have P–H bonds? H₃PO₂, H₃PO₃, H₃PO₄, H₄P₂O₇, H₄P₂O₅

   • A. 1
   • B. 2
   • C. 3
   • D. 4
   Answer and solution

   Answer: C

   H₃PO₂ has 2 P–H bonds, H₃PO₃ has 1 P–H bond, H₄P₂O₅ (hypophosphoric acid) has 2 P–H bonds (one per P). H₃PO₄ and H₄P₂O₇ have no P–H bonds. Total = 3 acids.

5. Q5 · Isolation of Metals

   In the extraction of copper from copper pyrite (CuFeS₂), the ore is roasted in a reverberatory furnace. During this roasting process, the major reaction occurring is:

   • A. 2CuFeS₂ + O₂ → Cu₂S + 2FeS + SO₂
   • B. 2FeS + 3O₂ → 2FeO + 2SO₂
   • C. 2CuFeS₂ + 4O₂ → Cu₂S + 3SO₂ + 2FeO
   • D. CuFeS₂ + 2O₂ → CuO + FeO + 2SO₂
   Answer and solution

   Answer: C

   During roasting in limited air supply, CuFeS₂ is converted to Cu₂S (matte) and iron is oxidized to FeO forming slag with SiO₂, while SO₂ is released.

6. Q6 · Alkenes and Alkynes

   An alkene (molecular formula C₆H₁₂) on ozonolysis followed by reductive work-up gives equimolar amounts of propanal and propanone. The alkene is:

   • A. 2-methylpent-2-ene
   • B. 2,3-dimethylbut-2-ene
   • C. Hex-3-ene
   • D. 2-methylpent-1-ene
   Answer and solution

   Answer: A

   The products CH₃CH₂CHO (propanal) and CH₃COCH₃ (propanone) indicate the double bond was between CH₃CH₂CH= and =C(CH₃)₂, which is 2-methylpent-2-ene.

7. Q7 · Alkyl Halides

   When 2-bromo-3-methylbutane is treated with alcoholic KOH, the major product formed follows Saytzeff's rule. The major alkene product is:

   • A. 2-Methylbut-1-ene
   • B. 2-Methylbut-2-ene
   • C. 3-Methylbut-1-ene
   • D. Pent-2-ene
   Answer and solution

   Answer: B

   Elimination of HBr from 2-bromo-3-methylbutane by alcoholic KOH proceeds via E2 mechanism. According to Saytzeff's rule, the more substituted alkene (2-methylbut-2-ene) is the major product.

8. Q8 · Amines

   Among the following, the correct order of basicity in aqueous solution is:

   • A. (CH₃)₂NH > CH₃NH₂ > NH₃ > C₆H₅NH₂
   • B. CH₃NH₂ > (CH₃)₂NH > NH₃ > C₆H₅NH₂
   • C. (CH₃)₂NH > CH₃NH₂ > C₆H₅NH₂ > NH₃
   • D. C₆H₅NH₂ > (CH₃)₂NH > CH₃NH₂ > NH₃
   Answer and solution

   Answer: A

   In aqueous solution, (CH₃)₂NH is most basic due to +I effect and optimal solvation. C₆H₅NH₂ is least basic due to resonance delocalization of the lone pair into the benzene ring.

9. Q9 · Polymers

   The monomer Y involved in the synthesis of Nylon 6 gives positive carbylamine test. If 8 moles of Y are analyzed using the Dumas method, the volume (in litres) of nitrogen gas evolved at STP is:

   • A. 89.6
   • B. 179.2
   • C. 44.8
   • D. 134.4
   Answer and solution

   Answer: A

   Nylon 6 is made from caprolactam, derived from ε-caprolactam containing one NH₂ per monomer. In Dumas method, 1 mol of N yields 1 mol N₂. 8 moles × 0.5 = 4 mol N₂. Volume = 4 × 22.4 = 89.6 L.

10. Q10 · Analytical Geometry — Two Dimensions

    Let E be an ellipse with foci F₁ and F₂. A point P on the ellipse satisfies ∠F₁PF₂ = π/3. If the semi-major axis is 4 and the semi-minor axis is 2√3, then the area of triangle F₁PF₂ is

    • A. 4√3
    • B. 6√3
    • C. 8√3
    • D. 12√3
    Answer and solution

    Answer: A

    For ellipse a = 4, b = 2√3, so c = √(16 - 12) = 2. Using |PF₁| + |PF₂| = 8 and the cosine rule with |F₁F₂| = 4 and angle π/3: |PF₁|² + |PF₂|² - 2|PF₁||PF₂|cos(π/3) = 16. Solving gives |PF₁||PF₂| = 16. Area = (1/2)|PF₁||PF₂|sin(π/3) = (1/2)(16)(√3/2) = 4√3.