A) \[mg\,\sin \theta +qB\sqrt{2g\,R\,\sin \,\theta }\]
B) \[3\,mg\,\sin \theta +qB\sqrt{2g\,R\,\sin \,\theta }\]
C) \[mg\,\sin \theta -qB\sqrt{2g\,R\,\sin \,\theta }\]
D) \[3\,mg\,\sin \theta -qB\sqrt{2g\,R\,\sin \,\theta }\]
Correct Answer: B
Solution :
[b] \[v=\sqrt{2gR\,\sin \theta }\] and \[F=qvB\] Now \[N-(F+mg\,\sin \theta )=\frac{m{{v}^{2}}}{R}\] \[\therefore \,\,N=F+mg\,\sin \,\theta +\frac{m{{v}^{2}}}{R}\] \[=qB\,\,\sqrt{2gR\,\sin \,\theta }+mg\sin \theta +2mg\sin \theta \] \[\therefore \,\,\,N=3mg\sin \theta +qB\sqrt{2gR\sin \theta }\]You need to login to perform this action.
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