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JEE Main 2023 · January session

A full paper in the 2023 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

19/52/19easy / medium / hard

Start timed mock Study 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

Units and measurement1
Kinematics1
Rotational motion1
Gravitation1
Simple harmonic motion1
Standing waves on strings1
Thermodynamics1
Kinetic theory of gases1
Capacitors1
Current electricity1
Moving charges in magnetic field1
Electromagnetic induction1
Alternating current1
Ray optics1

Chemistry · 30 questions

Chemical kinetics2
Atomic structure2
Chemical bonding2
Coordination compounds2
Mole concept1
Chemical equilibrium1
Chemical thermodynamics1
Electrochemistry1
Solutions1
Periodic properties1
p-Block elements1
d-Block elements1
General organic chemistry1
Hydrocarbons1

Mathematics · 30 questions

Matrices and determinants3
Complex numbers2
Sequences and series2
Permutations and combinations2
Binomial theorem2
Integration2
Three-dimensional geometry2
Probability2
Sets and relations1
Quadratic equations1
Limits1
Trigonometry1
Differentiation1
Applications of derivatives1

Sample question from this paper

Physics · Simple harmonic motion · medium

A block of mass

M

performs SHM of amplitude

A

on a smooth horizontal surface, attached to a spring. At the instant it passes through the mean position, a lump of putty of mass

M/3

is dropped on it from directly above and sticks. The new amplitude of oscillation is

A. $\dfrac{3}{4}A$
B. $A$
C. $\dfrac{2}{\sqrt3}A$
D. $\dfrac{\sqrt3}{2}A$

Show answer and solution

Answer: D
The putty has no horizontal momentum, so horizontal momentum is conserved at the mean position:

Mv=\tfrac43Mv'

, so

v'=\tfrac34v

with

v=\omega A

. The new angular frequency is

\omega'=\sqrt{k/(4M/3)}=\omega\sqrt{3/4}

. At the mean position speed is maximum, so

A'=\dfrac{v'}{\omega'}=\dfrac{\tfrac34\omega A}{\tfrac{\sqrt3}{2}\omega}=\dfrac{\sqrt3}{2}A

. (Equivalent: energy falls to

\tfrac34

, and

A'\propto\sqrt{E'}

at fixed

k

gives the same.)

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