A) \[\tan \theta =\frac{Rg}{{{v}^{2}}}\]
B) \[\tan \theta ={{v}^{2}}Rg\]
C) \[\tan \theta =\frac{{{v}^{2}}g}{R}\]
D) \[\tan \theta =\frac{{{v}^{2}}}{Rg}\]
Correct Answer: D
Solution :
\[R\,cos\theta =mg\] ...(i) \[R\sin \theta =\frac{m{{v}^{2}}}{R}\] ...(ii) From Eqs. (i) and (ii), we get \[\therefore \] \[\tan \theta =\frac{{{v}^{2}}}{Rg}\]You need to login to perform this action.
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