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JEE Main 2022 · June session

A full paper in the 2022 pattern: Physics, Chemistry and Mathematics, 20 MCQs + 10 numerical (answer any 5) each, one 180-minute clock. Take it as a timed mock or open it in learn mode and check every answer.

90questions, 75 counted
180 minone clock
300maximum marks
17/58/15easy / medium / hard
Start timed mockStudy with solutions

Marking scheme and topics

+4 for a correct answer, −1 for a wrong MCQ or numerical answer. Each subject has 30 questions: Section A multiple choice, Section B numerical value.

Physics · 30 questions

  • Capacitors and dielectrics2
  • Units and measurements1
  • Projectile motion1
  • Laws of motion and friction1
  • Circular motion and energy1
  • Rotational motion1
  • Gravitation1
  • Simple harmonic motion1
  • Waves and Doppler effect1
  • Thermodynamics1
  • Heat transfer1
  • Gauss's law1
  • Current electricity1
  • Moving charges in magnetic field1

Chemistry · 30 questions

  • Electrochemistry2
  • Chemical kinetics2
  • Solutions2
  • Coordination compounds2
  • p-block elements2
  • Mole concept1
  • Chemical equilibrium1
  • Atomic structure1
  • Thermodynamics1
  • General organic chemistry1
  • Alkenes and carbonyl compounds1
  • Haloalkanes1
  • Alcohols1
  • Aldehydes and ketones1

Mathematics · 30 questions

  • Permutations and combinations2
  • Binomial theorem2
  • Matrices and determinants2
  • Definite integrals2
  • Three-dimensional geometry2
  • Vector algebra2
  • Probability2
  • Sequences and series2
  • Complex numbers1
  • Quadratic equations1
  • Limits1
  • Differentiation1
  • Application of derivatives1
  • Indefinite integrals1

Sample question from this paper

Physics · Laws of motion and friction · medium

A block of mass 5 kg5\ \mathrm{kg} rests on a rough plane inclined at 53∘53^\circ to the horizontal; the coefficient of static friction is 0.50.5. The minimum horizontal force that must be applied to the block (pushing it towards the incline) to keep it from sliding down is (g=10 m s−2g=10\ \mathrm{m\,s^{-2}}, sin⁡53∘=0.8\sin53^\circ=0.8)
  1. A. 15 N15\ \mathrm{N}
  2. B. 25 N25\ \mathrm{N}
  3. C. 30 N30\ \mathrm{N}
  4. D. 40 N40\ \mathrm{N}
Show answer and solution
Answer: B
Resolve along and perpendicular to the incline. The horizontal force FF has components Fcos⁡θF\cos\theta up the incline and Fsin⁡θF\sin\theta into the plane. Normal reaction N=mgcos⁡θ+Fsin⁡θN=mg\cos\theta+F\sin\theta. At the verge of sliding down: Fcos⁡θ+μN=mgsin⁡θF\cos\theta+\mu N=mg\sin\theta, so F(cos⁡θ+μsin⁡θ)=mg(sin⁡θ−μcos⁡θ)F(\cos\theta+\mu\sin\theta)=mg(\sin\theta-\mu\cos\theta). With cos⁡53∘=0.6\cos53^\circ=0.6: F(0.6+0.4)=50(0.8−0.3)F(0.6+0.4)=50(0.8-0.3), giving F=25 NF=25\ \mathrm{N}. Forgetting the extra normal force from Fsin⁡θF\sin\theta gives 41.7 N41.7\ \mathrm{N}.

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