A) \[\sqrt{g{{R}^{2}}\frac{{{\mu }_{s}}+\tan \theta }{1-{{\mu }_{s}}\tan \theta }}\]
B) \[\sqrt{gR\frac{{{\mu }_{s}}+\tan \theta }{1-{{\mu }_{s}}\tan \theta }}\]
C) \[\sqrt{\frac{g}{R}\frac{{{\mu }_{s}}+\tan \theta }{1-{{\mu }_{s}}\tan \theta }}\]
D) \[\sqrt{\frac{g}{{{R}^{2}}}\frac{{{\mu }_{s}}+\tan \theta }{1-{{\mu }_{s}}\tan \theta }}\]
Correct Answer: B
Solution :
[b] On a banked road, \[\frac{{{v}^{2}}\max }{Rg}=\left( \frac{{{\mu }_{s}}+\tan \theta }{1-{{\mu }_{s}}\tan \theta } \right)\] Maximum safe velocity of a car on the banked road \[{{V}_{\max }}=\sqrt{Rg\left[ \frac{{{\mu }_{s}}+\tan \theta }{1-{{\mu }_{s}}\tan \,\theta } \right]}\]
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