JEE Main 2019PhysicsLaws of MotionMediumMCQ

JEE Main 2019 — Laws of Motion Question with Solution

JEE Main 2019 (10 Apr Shift 2)

Question

Two blocks  A and  B of masses mA=1 kg and mB=3 kg are kept on the table as shown in figure. The coefficients of friction between A and B is 0.2 and between  B and the surface of the table is also 0.2. The maximum force F that can be applied on B horizontally, so that the block  A does not slide over the block B is : [Take  g=10 m/s2 ]

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Show full solutionCorrect option: A
Correct answer
A16 N

Step-by-step explanation

Maximum possible acceleration of 1 kg block a=µmgm=µg=2 m/s2

For 1 g block not to slide the maximum acceleration of both blocks should be

Fmax-µmg=mamax
Fmax-0.2×4×10=4×2
Fmax- 8=8
Fmax=16 N

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About this question

This is a previous-year question from JEE Main 2019, covering the Laws of Motion chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.