JEE Mains · Physics · STD 11 - 4.2 friction
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 \(............\,cm\)
- A \(2\)
- B \(1\)
- C \(8\)
- D \(4\)
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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