JEE Main 2021 — Laws Of Motion Question with Solution
From: JEE Main 2021 (Online) 17th March Morning Shift
Question
Two blocks (m = 0.5 kg and M = 4.5 kg) are arranged on a horizontal frictionless table as shown in figure. The coefficient of static friction between the two blocks is . 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 ms2]
This question includes a diagram. The text above accompanies the figure.
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Show full solutionCorrect answer: 21
Correct answer
21
Step-by-step explanation
f = m = m
m mg (for no slipping)
F (m + M)g
Fmax = = 21 N
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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.