# A cyclist riding the bicycle at a speed of 14√3 m/s takes a turn around a circular road of radius 20√3 m without skidding. Given, g = 9.8 ms^-2 what is his inclination to the vertical?

Q: A cyclist riding the bicycle at a speed of 14√3 m/s takes a turn around a circular road of radius 20√3 m without skidding. Given, g = 9.8 ms-2 what is his inclination to the vertical?

(a) 30°

(b) 90°

(c) 45°

(d) 60°

$tan\theta = \frac{v^2}{r g}$
$tan\theta = \frac{(14\sqrt{3})^2}{20\sqrt{3}\times 9.8}$
$tan\theta = \sqrt{3}$
$\theta = 60^o$