A) 172 m/s
B) 124 m/s
C) 99 m/s
D) 86 m/s
Correct Answer: A
Solution :
The maximum velocity for a banked road with friction, \[{{v}^{2}}=gr\left( \frac{\mu +\tan \theta }{1-\mu \tan \theta } \right)\] Þ \[{{v}^{2}}=9.8\times 1000\times \left( \frac{0.5+1}{1-0.5\times 1} \right)\]Þ \[v=172\,m/s\]
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