JEE Main 2019PhysicsLaws of MotionEasyMCQ

JEE Main 2019 — Laws of Motion Question with Solution

JEE Main 2019 (09 Jan Shift 2)

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 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is g=10 m s-2

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Show full solutionCorrect option: A
Correct answer
A100 N

Step-by-step explanation


Tcos⁡45°=mg

Tsin⁡45°=F

tan 45°=Fmg

F=mg=100 N

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About this question

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.