Question

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

• A. ${45}^{\circ }$
• B. ${40}^{\circ }$
• C. ${42}^{\circ }$
• D. ${48}^{\circ }$

Answer 1

The correct option is A

${45}^{\circ }$

$2\pi r=34.3\Rightarrow r=\frac{34.3}{2\pi }andv=\frac{2\pi r}{T}=\frac{2\pi r}{\sqrt{22}}$
Anglle of binding $\theta =ta{n}^{-1}\left(\frac{v^2}{rg}\right)={45}^{\circ }$