JEE Main 2020PhysicsLaws of MotionMediumMCQ

JEE Main 2020 — Laws of Motion Question with Solution

JEE Main 2020 (07 Jan Shift 2)

Question

A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of 45° with the vertical. Then F equals: (Take g=10 m s-2 and the rope to be massless)

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Show full solutionCorrect option: A
Correct answer
A100N

Step-by-step explanation

Let the tension in the string is T.

Applying the condition of equilibrium in the vertical direction,

Tcos45°=100

⇒ T2=100   ...1

In the horizontal direction,

Tsin45°=F

⇒ T2=F

Put the value of T from equation 1,

⇒ F=100 N

So, the horizontal force applied on the rope will be 100 N.

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About this question

This is a previous-year question from JEE Main 2020, 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.