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4-1.Newton's Laws of Motion
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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 ....... $^o$
A$45$
B$40$
C$42$
D$48$
Solution
(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 $
Angle of binding $\theta = {\tan ^{ – 1}}\left( {\frac{{{v^2}}}{{rg}}} \right) = 45^\circ $
Standard 11
Physics
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