A cyclist goes round a circular path of circumference $34.3\, m$ in $\sqrt {22} $ sec. the angle made by him, with the vertical, will be ....... $^o$
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(a) $2\pi r = 34.3$ $⇒$ $r = \frac{{34.3}}{{2\pi }}$ and $v = \frac{{2\pi r}}{T} = \frac{{2\pi r}}{{\sqrt {22} }}$
Angle of binding $\theta = {\tan ^{ - 1}}\left( {\frac{{{v^2}}}{{rg}}} \right) = 45^\circ $
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