KVPY Sample Paper KVPY Stream-SX Model Paper-22

  • question_answer
    A charged sphere of mass m and charge - q starts sliding along the surface of a smooth hemispherical bowl, at position\[P\]. The region has a transverse uniform magnetic field \[B\]. Normal force by the surface of bowl on the sphere at position \[Q\]is

    A) \[mg\sin \theta +qB\sqrt{2gR\sin \theta }\]

    B) \[3\,mg\sin \theta +qB\sqrt{2gR\sin \theta }\]

    C) \[\,mg\sin \theta -qB\sqrt{2gR\sin \theta }\]

    D) \[3\,mg\sin \theta -qB\sqrt{2gR\sin \theta }\]

    Correct Answer: B

    Solution :

                            \[v=\sqrt{2gR\sin \theta }\]
    and                   \[F=qvB\]
    Now \[N-(F+\operatorname{mg}\,sin\theta )=\frac{m{{v}^{2}}}{R}\]
    \[\therefore \]\[N=F+mg\sin \theta +\frac{m{{v}^{2}}}{R}\]
    \[a=qB\sqrt{2gR\sin \theta }+mg\sin \theta +2mg\sin \theta \]
    \[\therefore \]\[N=3mg\sin \theta +qB\sqrt{2gR\sin \theta }\]

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