Q: A cyclist riding at a speed of 9.8 m/s takes a turn around a circular road of radius 19.6 m. What is his inclination to the vertical?

(a) 30°

(b) 45°

(c) 10.5°

(d) 26.5°

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Ans: (d)

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

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

$tan\theta = \frac{9.8^2}{19.6 \times 9.8} $

$tan\theta = \frac{1}{2} $

$\theta = 26.5^o$