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.
You will be redirected in
3 sec