JEE Main 2020 — Laws Of Motion Question with Solution
From: JEE Main 2020 (Online) 7th January Evening Slot
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 45o
with the vertical. Then F equal : (Take g = 10 ms–2 and the rope to be massless)
Choose an option
Show full solutionCorrect option: A
Correct answer
A100 N
Step-by-step explanation
For equilibrium,
T sin 45o = F ....(1)
and T cos 45o = 10g ....(2)
Performing (1)/(2)
we get F = 10g = 100 N
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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.