JEE Main 2021PhysicsLaws of MotionEasyNumerical

JEE Main 2021Laws of Motion Question with Solution

JEE Main 2021 (17 Mar Shift 1)

Question

Two blocks (m=0.5 kg and M=4.5 kg) are arranged on a horizontal frictionless table as shown in the figure. The coefficient of static friction between the two blocks is 37. Then the maximum horizontal force that can be applied on the larger block so that the blocks move together is N. (Round off to the Nearest Integer) [Take g as 9.8 m s-2]

Enter your answer

Show full solutionCorrect answer: 21
Correct answer
21

Step-by-step explanation

amax=μg=37×9.8

F=(M+m)amax=5amax

=21 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 2021, 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.