JEE Main 2022PhysicsLaws of MotionEasyMCQ

JEE Main 2022 — Laws of Motion Question with Solution

JEE Main 2022 (27 Jun Shift 1)

Question

A system of two blocks of masses m=2 kg and M=8 kg is placed on a smooth table as shown in figure. The coefficient of static friction between two blocks is 0.5. The maximum horizontal force F that can be applied to the block of mass M so that the blocks move together will be g=9.8 m s-2

Choose an option

Show full solutionCorrect option: C
Correct answer
C49 N

Step-by-step explanation

Maximum acceleration of the upper block will be,

amax=μg=0.5×9.8=4.9 m s-2

Therefore, maximum force that can be applied on the lower block is,

Fmax=2+8×4.9=49 N

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Laws of Motion chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2022, 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.