JEE Main 2018 — Circular Motion Question with Solution
From: JEE Main 2018 (Online) 15th April Evening Slot
Question
A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of revolutions per second. A coin placed at a distnce of 1.25 cm from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is : (g = 10 m/s2)
Choose an option
Show full solutionCorrect option: D
Correct answer
D0.6
Step-by-step explanation
We have
...... (1)
Given : rate of rotation = 3.5 rev/s
1 revolution = 2 rad
That is, 3.5 revolutions = 3.5 2 rad
Therefore, = 3.5 2 rad/s
r = 1.25 cm = 1.25 102 m
Thus, from Eq. (1), we have
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This is a previous-year question from JEE Main 2018, covering the Circular 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.