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NEET Sample Paper NEET Sample Test Paper-17

  • question_answer
    A hollow cylinder drum or radius r is placed with its axis vertical. It is rotated about an axis passing through its centre and perpendicular to the base and a body is placed on the inside surface of the drum. If the coefficient of friction is , the minimum frequency of revolution so that the body does not fall down is:

    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].


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