A cyclist riding the bicycle at a speed of $14 \sqrt{3} \,m / s$ takes a turn around a circular road of radius $20 \sqrt{3} \,m$ without skidding. What is his inclination to the vertical?
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(c)
$\tan \theta=\frac{v^2}{r g}$
$=\frac{14 \times 14 \times 3}{20 \sqrt{3} \times 10} \simeq \sqrt{3}$
$\theta=60^{\circ}$
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