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JEE Main 2025 · January session

A full paper in the 2025 pattern: Physics, Chemistry and Mathematics, 20 MCQs + 5 numerical each, one 180-minute clock. Take it as a timed mock or open it in learn mode and check every answer.

75questions
180 minone clock
300maximum marks
14/44/17easy / 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 25 questions: Section A multiple choice, Section B numerical value.

Physics · 25 questions

  • Units and measurement1
  • Laws of motion and friction1
  • Circular motion and energy1
  • Rotational motion1
  • Gravitation1
  • Fluid mechanics1
  • Thermodynamics1
  • Simple harmonic motion1
  • Waves and Doppler effect1
  • Electrostatics1
  • Semiconductor devices1
  • Current electricity1
  • Moving charges in magnetic field1
  • Magnetic effect of current1

Chemistry · 25 questions

  • Solutions and colligative properties2
  • Hydrocarbons2
  • Amines and diazonium salts2
  • Coordination compounds2
  • Mole concept1
  • Atomic structure1
  • Chemical equilibrium1
  • Thermodynamics1
  • Electrochemistry1
  • Chemical kinetics1
  • General organic chemistry1
  • Stereochemistry1
  • Biomolecules1
  • Alcohols, phenols and ethers1

Mathematics · 25 questions

  • Permutations and combinations2
  • Definite integration2
  • Circles2
  • Sets and relations1
  • Complex numbers1
  • Quadratic equations1
  • Statistics1
  • Binomial theorem1
  • Matrices1
  • Determinants1
  • Limits1
  • Continuity and differentiability1
  • Differentiation1
  • Application of derivatives1

Sample question from this paper

Physics · Laws of motion and friction · medium

A small block slides from rest down a rough incline of angle 30∘30^\circ and takes 1.51.5 times as long to reach the bottom as it would on a perfectly smooth incline of the same angle and length. The coefficient of kinetic friction between the block and the incline is
  1. A. 133\dfrac{1}{3\sqrt{3}}
  2. B. 593\dfrac{5}{9\sqrt{3}}
  3. C. 59\dfrac{5}{9}
  4. D. 123\dfrac{1}{2\sqrt{3}}
Show answer and solution
Answer: B
For the same distance ss from rest, s=12at2s=\tfrac12 a t^2, so t∝a−1/2t\propto a^{-1/2} and arough=asmooth/n2a_{\text{rough}} = a_{\text{smooth}}/n^2 with n=1.5n = 1.5. Smooth: as=gsin⁡θa_s = g\sin\theta. Rough: ar=gsin⁡θ−μgcos⁡θa_r = g\sin\theta - \mu g\cos\theta. Then gsin⁡θ−μgcos⁡θ=gsin⁡θ/2.25g\sin\theta-\mu g\cos\theta = g\sin\theta/2.25, giving μcos⁡θ=sin⁡θ(1−49)\mu\cos\theta = \sin\theta\left(1-\tfrac{4}{9}\right), i.e. μ=59tan⁡30∘=593≈0.32\mu = \tfrac59\tan 30^\circ = \dfrac{5}{9\sqrt3}\approx 0.32. Using nn instead of n2n^2 gives the wrong value 133\tfrac{1}{3\sqrt3}.

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