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JEE Main 2016 · April (offline)

A full paper in the 2016 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
18/50/22easy / 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

  • Kinematics1
  • Circular motion1
  • Friction and relative motion1
  • Power1
  • Conservation of angular momentum1
  • Gravitation1
  • Surface tension1
  • Calorimetry1
  • Thermodynamics1
  • Kinetic theory of gases1
  • Standing waves in pipes1
  • Waves on a string1
  • Capacitors with dielectrics1
  • Electrostatics of conductors1

Chemistry · 30 questions

  • Electrochemistry2
  • Mole concept1
  • Chemical equilibrium1
  • Thermochemistry1
  • Chemical kinetics1
  • Solutions1
  • Atomic structure1
  • Ionic equilibrium1
  • Solid state1
  • General organic chemistry1
  • Hydrocarbons1
  • Haloalkanes1
  • Aldehydes and ketones1
  • Amines1

Mathematics · 30 questions

  • Complex numbers2
  • Quadratic equations2
  • Sequences and series2
  • Permutations and combinations2
  • Determinants2
  • Applications of derivatives2
  • Sets1
  • Relations1
  • Binomial theorem1
  • Trigonometry1
  • Matrices1
  • Limits1
  • Continuity1
  • Differentiation1

Sample question from this paper

Physics · Circular motion · medium

A small bead slides without friction on a smooth circular wire hoop of radius 0.5 m0.5\ \mathrm{m} kept in a vertical plane. The hoop rotates uniformly about its vertical diameter. Taking g=10 m s−2g=10\ \mathrm{m\,s^{-2}}, the angular speed of the hoop for which the bead stays at rest relative to the hoop at a position where its radius makes an angle of 60∘60^\circ with the downward vertical is
  1. A. 20 rad s−1\sqrt{20}\ \mathrm{rad\,s^{-1}}
  2. B. 210 rad s−12\sqrt{10}\ \mathrm{rad\,s^{-1}}
  3. C. 10 rad s−110\ \mathrm{rad\,s^{-1}}
  4. D. 4.8 rad s−14.8\ \mathrm{rad\,s^{-1}}
Show answer and solution
Answer: B
The bead moves in a horizontal circle of radius r=Rsin⁡θr=R\sin\theta with θ=60∘\theta=60^\circ. The only forces are gravity and the normal reaction NN directed along the radius towards the centre of the hoop. Vertical: Ncos⁡θ=mgN\cos\theta=mg. Horizontal: Nsin⁡θ=mω2Rsin⁡θN\sin\theta=m\omega^2R\sin\theta, which gives N=mω2RN=m\omega^2R. Combining, cos⁡θ=gω2R\cos\theta=\dfrac{g}{\omega^2R}, so ω2=gRcos⁡60∘=100.5×0.5=40\omega^2=\dfrac{g}{R\cos60^\circ}=\dfrac{10}{0.5\times0.5}=40 and ω=210≈6.3 rad s−1\omega=2\sqrt{10}\approx6.3\ \mathrm{rad\,s^{-1}}. Setting ω2=g/R=20\omega^2=g/R=20 would hold only at θ=0\theta=0 or cos⁡θ=1\cos\theta=1, so the answer must be larger than 20\sqrt{20}.

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