A) \[\frac{mg\,R\,\sin \theta }{{{B}^{2}}{{l}^{2}}}\]
B) \[\frac{mg\,R\,\sin \theta }{B{{l}^{2}}}\]
C) \[\frac{mg\,R\,\sin \theta }{{{B}^{2}}{{l}^{5}}}\]
D) \[\frac{mg\,R\,\sin \theta }{B{{l}^{4}}}\]
Correct Answer: A
Solution :
[a] Component of weight along the inclied plane \[=mg\sin \theta \] Again, \[F=BI\ell =B\frac{B\ell v}{R}\ell =\frac{{{B}^{2}}{{\ell }^{2}}v}{R}\] Now, \[\frac{{{B}^{2}}{{\ell }^{2}}v}{R}=mg\,\sin \theta \] or \[v=\frac{mgR\,\sin \,\theta }{{{B}^{2}}{{\ell }^{2}}}\]You need to login to perform this action.
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