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JEE Main 2024 · January session

A full paper in the 2024 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
18/55/17easy / 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

  • Rotational motion2
  • Units and errors1
  • Relative motion1
  • Projectile motion1
  • Constraint motion1
  • Work, energy and power1
  • Gravitation1
  • Surface tension1
  • Thermodynamics1
  • Kinetic theory of gases1
  • Waves and sound1
  • Electrostatics1
  • Electric field and oscillations1
  • Current electricity1

Chemistry · 30 questions

  • Electrochemistry2
  • Coordination compounds2
  • Amines2
  • Mole concept1
  • Chemical equilibrium1
  • Thermodynamics (calorimetry)1
  • Chemical kinetics1
  • Solutions and colligative properties1
  • Atomic structure1
  • p-Block elements1
  • d- and f-block elements1
  • Chemical bonding (VSEPR)1
  • Metallurgy1
  • Qualitative analysis1

Mathematics · 30 questions

  • Sets and relations2
  • Sequences and series2
  • Binomial theorem2
  • Matrices and determinants2
  • Definite integration2
  • Differential equations2
  • Vectors2
  • Probability2
  • Complex numbers1
  • Quadratic equations1
  • Permutations and combinations1
  • Limits1
  • Continuity1
  • Differentiation1

Sample question from this paper

Physics · Relative motion · medium

A boat can move at 5 m s−15\ \mathrm{m\,s^{-1}} relative to still water. It crosses a river 120 m120\ \mathrm{m} wide that flows at 3 m s−13\ \mathrm{m\,s^{-1}}. The crossing time when the boat is steered to land at the point directly opposite the start, exceeds the minimum possible crossing time by
  1. A. 6 s6\ \mathrm{s}
  2. B. 8 s8\ \mathrm{s}
  3. C. 10 s10\ \mathrm{s}
  4. D. 4 s4\ \mathrm{s}
Show answer and solution
Answer: A
Minimum time: head perpendicular to the banks, tmin=120/5=24 st_{min}=120/5=24\ \mathrm{s}. To land directly opposite, the boat must cancel the drift: 5sin⁡θ=35\sin\theta=3, so the cross-stream component of velocity is 52−32=4 m s−1\sqrt{5^2-3^2}=4\ \mathrm{m\,s^{-1}} and t=120/4=30 st=120/4=30\ \mathrm{s}. Difference =30−24=6 s=30-24=6\ \mathrm{s}. A common slip is using 5−3=25-3=2 or 5+35+3 as the cross velocity.

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