A) \[{{\tan }^{-1}}\left( \frac{{{60}^{2}}}{0.1} \right)\]
B) \[{{\tan }^{-1}}\left( \frac{{{(50/2)}^{2}}}{100\times 9.8} \right)\]
C) \[{{\tan }^{-1}}{{(60\times 0.1\times 9.8)}^{{1}/{2}\;}}\]
D) \[{{\tan }^{-1}}\left( \frac{100\times 9.8}{{{(50/3)}^{3}}} \right)\]
Correct Answer: B
Solution :
For a body moving with velocity v on a circular track of radius R, the angle of banking is given by \[\tan \theta =\frac{{{v}^{2}}}{Rg}\] Given, \[v=60\text{ }km/h=\frac{60\times 1000}{60\times 60}=\frac{50}{3}m/s\] \[R=0.1\text{ }km=100m\] \[\tan \theta =\frac{{{\left( \frac{50}{3} \right)}^{2}}}{100\times 9.8}\] \[\Rightarrow \] \[\theta ={{\tan }^{-1}}\left( \frac{{{(50/3)}^{2}}}{100\times 9.8} \right)\]
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