A) \[\frac{1}{2\pi }\sqrt{\frac{\mu g}{r}}\]
B) \[\frac{1}{2\pi }\sqrt{\frac{g}{\mu r}}\]
C) \[\frac{1}{2\pi }\sqrt{\frac{g}{2\mu r}}\]
D) \[\frac{1}{2\pi }\sqrt{\frac{2\mu g}{r}}\]
Correct Answer: B
Solution :
The body will not fall down till, force of friction \[\ge \] weight of body \[F\ge mg\] \[\mu F\ge mg\] \[\mu (mr{{\omega }^{2}})\ge mg\] \[\omega \ge \sqrt{\frac{g}{\mu r}}\] \[2\pi n\ge \sqrt{\frac{g}{\mu r}}\] \[\therefore \]Minimum value of \[n=\frac{1}{2\pi }\sqrt{\frac{g}{\mu r}}\] Hence, the correction option is [b].You need to login to perform this action.
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