A) \[\frac{1}{2\pi }\sqrt{\frac{\mu g}{a}}\]
B) \[\sqrt{\frac{\mu g}{a}}\]
C) \[2\pi \sqrt{\frac{a}{\mu g}}\]
D) \[\sqrt{\frac{a}{\mu g}}\]
Correct Answer: A
Solution :
[a] For the block is about to slip, \[\mu g={{\omega }^{2}}a\] \[\omega =\sqrt{\frac{\mu g}{a}}\Rightarrow v=\frac{1}{2\pi }\sqrt{\frac{\mu g}{a}}\]
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