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°

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Ans: (d)

$tan\theta = \frac{v^2}{r g} $

$tan\theta = \frac{(14\sqrt{3})^2}{20\sqrt{3}\times 9.8} $

$tan\theta = \sqrt{3}$

$\theta = 60^o$