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JEE Main 2019 · January session

A full paper in the 2019 pattern: Physics, Chemistry and Mathematics, 30 MCQs each, one 180-minute clock. Take it as a timed mock or open it in learn mode and check every answer.

90questions
180 minone clock
360maximum marks
23/49/18easy / medium / hard
Start timed mockStudy with solutions

Marking scheme and topics

+4 for a correct answer, −1 for a wrong answer. Each subject has 30 questions.

Physics · 30 questions

  • Heat transfer2
  • Units and errors1
  • Kinematics1
  • Kinematics and collisions1
  • Circular motion1
  • Work, energy and power1
  • Angular momentum1
  • Gravitation1
  • Fluid mechanics1
  • Thermodynamics1
  • Kinetic theory of gases1
  • Oscillations1
  • Waves on a string1
  • Doppler effect1

Chemistry · 30 questions

  • Electrochemistry2
  • Solutions and colligative properties2
  • Chemical bonding2
  • Coordination compounds2
  • Mole concept1
  • Chemical equilibrium1
  • Chemical thermodynamics1
  • Chemical kinetics1
  • Atomic structure1
  • Ionic equilibrium1
  • General organic chemistry1
  • Hydrocarbons1
  • Haloalkanes1
  • Alcohols, phenols and ethers1

Mathematics · 30 questions

  • Matrices and Determinants3
  • Complex Numbers2
  • Quadratic Equations2
  • Probability2
  • Binomial Theorem2
  • Definite Integrals2
  • Differential Equations2
  • Sets1
  • Relations and Functions1
  • Sequences and Series1
  • Permutations and Combinations1
  • Limits1
  • Trigonometry1
  • Continuity and Differentiability1

Sample question from this paper

Physics · Circular motion · medium

A car travels on a rough circular road of radius 45 m45\ \mathrm{m} that is banked at an angle θ\theta with tan⁡θ=0.75\tan\theta=0.75. The coefficient of friction between the tyres and the road is 0.50.5. Take g=10 m s−2g=10\ \mathrm{m\,s^{-2}}. The maximum speed with which the car can go round the curve without skidding outward is
  1. A. 15 m s−115\ \mathrm{m\,s^{-1}}
  2. B. 18.4 m s−118.4\ \mathrm{m\,s^{-1}}
  3. C. 30 m s−130\ \mathrm{m\,s^{-1}}
  4. D. 36 m s−136\ \mathrm{m\,s^{-1}}
Show answer and solution
Answer: C
At the limit, friction acts down the slope. Resolving along the horizontal (centripetal) and vertical directions: Nsin⁡θ+fcos⁡θ=mv2rN\sin\theta+f\cos\theta=\dfrac{mv^2}{r} and Ncos⁡θ−fsin⁡θ=mgN\cos\theta-f\sin\theta=mg with f=μNf=\mu N. Dividing, v2rg=tan⁡θ+μ1−μtan⁡θ=0.75+0.51−0.375=2\dfrac{v^2}{rg}=\dfrac{\tan\theta+\mu}{1-\mu\tan\theta}=\dfrac{0.75+0.5}{1-0.375}=2. So v2=2(45)(10)=900v^2=2(45)(10)=900 and v=30 m s−1v=30\ \mathrm{m\,s^{-1}}. Ignoring friction gives rgtan⁡θ≈18.4 m s−1\sqrt{rg\tan\theta}\approx18.4\ \mathrm{m\,s^{-1}}; a flat road would give μrg=15 m s−1\sqrt{\mu rg}=15\ \mathrm{m\,s^{-1}}.

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