JEE Main 2019 — Laws Of Motion Question with Solution
From: JEE Main 2019 (Online) 9th January Evening Slot
Question
A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point. the rope deviated at an angle of 45o at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms2
Choose an option
Show full solutionCorrect option: D
Correct answer
D100 N
Step-by-step explanation
tan 45o =
F = mg
= 10 10
= 100 N
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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.