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JEE Mains · Physics · STD 11 - Friction
  • JEE Mains 2023
  • Medium
  • Numerical

A coin placed on a rotating table just slips when it is placed at a distance of \(1\,cm\) from the center. If the angular velocity of the table in halved, it will just slip when placed at a distance of from the centre \(............\,cm\)

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Answer & Solution

Correct Answer

(D) \(4\)

Step-by-step Solution

Detailed explanation

\(f _{ s \max }=\mu mg = m \omega^2 R \Rightarrow R =\frac{\mu g }{\omega^2}\)

So if \(\omega\) becomes \(\frac{\omega}{2}, R\) will become \(4 R\).

So distance from the center will be \(4\,cm\).

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