JEE Main 2022 previous year paper practice · PYQ-pattern set

JEE Main 2022 PYQ-pattern questions

Showing 10 of the 15 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 · Laws of Motion

   A block of mass 4 kg is placed on a horizontal surface. A horizontal force of 20 N is applied on the block. If the coefficient of kinetic friction between the block and the surface is 0.3, what is the acceleration of the block? (Take g = 10 m/s²)

   • A. 2.0 m/s²
   • B. 3.5 m/s²
   • C. 5.0 m/s²
   • D. 2.5 m/s²
   Answer and solution

   Answer: A

   Frictional force f = μN = 0.3 × 4 × 10 = 12 N. Net force = 20 - 12 = 8 N. Acceleration a = F_net/m = 8/4 = 2.0 m/s².

2. Q2 · Thermodynamics

   An ideal gas undergoes a cyclic process ABCA where AB is isothermal expansion, BC is adiabatic expansion, and CA is constant volume process. If work done during AB is 500 J and during BC is 300 J, what is the heat rejected during CA?

   • A. 800 J
   • B. 500 J
   • C. 200 J
   • D. 300 J
   Answer and solution

   Answer: D

   In a cycle, net heat equals net work. Net work = W_AB + W_BC + W_CA = 500 + 300 + 0 = 800 J. Heat added: Q_AB = 500 J (isothermal), Q_BC = 0 (adiabatic). So Q_CA = net work − heat added = 800 − 500 = −300 J, meaning 300 J is rejected during CA.

3. Q3 · Current Electricity

   A wire of resistance R is stretched uniformly to twice its original length. The resistance of the stretched wire will be

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

   Answer: C

   When a wire is stretched to twice its length while keeping volume constant, its cross-sectional area becomes A/2. Resistance R = ρL/A, so new resistance = ρ(2L)/(A/2) = 4ρL/A = 4R.

4. Q4 · Optics

   A convex lens of focal length 20 cm forms a real image at a distance of 60 cm from the lens. The object distance (in cm) is

   • A. 15
   • B. 30
   • C. 40
   • D. 45
   Answer and solution

   Answer: B

   Using lens formula 1/f = 1/v - 1/u, where f = 20 cm and v = 60 cm. Thus 1/20 = 1/60 - 1/u, solving gives 1/u = -1/30, so u = -30 cm (object distance is 30 cm).

5. Q5 · Atoms and Nuclei

   A radioactive nucleus has a half-life of 30 days. What fraction of the sample will remain undecayed after 90 days?

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

   Answer: C

   Number of half-lives = 90/30 = 3. Remaining fraction = (1/2)³ = 1/8.

6. Q6 · Chemical Thermodynamics

   For a reaction at equilibrium, ΔG° = +5.7 kJ/mol at 298 K. The equilibrium constant K for this reaction is approximately: (R = 8.314 J/mol·K, ln 10 = 2.303)

   • A. 0.1
   • B. 1.0
   • C. 10
   • D. 100
   Answer and solution

   Answer: A

   ΔG° = -RT ln K. 5700 = -8.314 × 298 × ln K. ln K = -2.30. K = e⁻²·³⁰ ≈ 0.1.

7. Q7 · Chemical Kinetics

   A first-order reaction has a rate constant of 0.035 min⁻¹. How much time will it take for the concentration of the reactant to decrease to 25% of its initial value?

   • A. 39.6 min
   • B. 49.8 min
   • C. 19.8 min
   • D. 29.7 min
   Answer and solution

   Answer: A

   For first order: t = (2.303/k) log([A]₀/[A]). Here, [A] = 0.25[A]₀, so t = (2.303/0.035) log(4) = 65.8 × 0.602 = 39.6 min.

8. Q8 · Solutions

   What is the molality of a solution prepared by dissolving 18 g of glucose (C₆H₁₂O₆) in 500 g of water? (Molar mass of glucose = 180 g/mol)

   • A. 0.1 m
   • B. 0.2 m
   • C. 0.3 m
   • D. 0.4 m
   Answer and solution

   Answer: B

   Moles of glucose = 18/180 = 0.1 mol. Molality = moles/kg of solvent = 0.1/0.5 = 0.2 m.

9. Q9 · Permutations and Combinations

   The number of ways to arrange the letters of the word ENGINEERING so that all the three E's are together is:

   • A. 9! / (3! × 2! × 2!)
   • B. 9! / (2! × 2!)
   • C. 8! / (2! × 2!)
   • D. 11! / (3! × 3! × 2! × 2!)
   Answer and solution

   Answer: A

   ENGINEERING has E×3, N×3, G×2, I×2, R×1. Treating the three E's as a single block gives 9 items: block, N,N,N, G,G, I,I, R. Arrangements = 9!/(3! for N's × 2! for G's × 2! for I's). This matches option A.

10. Q10 · Vector Algebra

    If |a| = 3, |b| = 4, and |a × b| = 6, then a·b equals:

    • A. 6√3
    • B. 3√11
    • C. ±6√3
    • D. ±3√11
    Answer and solution

    Answer: C

    We know |a × b| = |a||b|sin θ, so 6 = 12 sin θ, giving sin θ = 1/2. Then cos θ = ±√3/2. Since a·b = |a||b|cos θ = 12(±√3/2) = ±6√3.