JEE Main 2021 — Laws Of Motion Question with Solution
From: JEE Main 2021 (Online) 1st September Evening Shift
Question
An object of mass 'm' is being moved with a constant velocity under the action of an applied force of 2N along a frictionless surface with following surface profile.
The correct applied force vs distance graph will be :
The correct applied force vs distance graph will be :
This question includes a diagram. The text above accompanies the figure.
Choose an option
Show full solutionCorrect option: C
Correct answer
C
Step-by-step explanation
During upward motion
F = 2N = (+ve) constant
During downward motion
F = 2N = (ve) constant
Best possible answer is option (c)
F = 2N = (+ve) constant
During downward motion
F = 2N = (ve) constant
Best possible answer is option (c)
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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.