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CGP EDU Academic Team
Published on: September 12, 2026
A cyclist riding the bicycle at a speed of \(14 \sqrt{3}\) ms –1 takes a turn around a circular road of radius \(20 \sqrt{3}\) m without skidding. Given g = 9.8 ms –2 , what is his inclination to the vertical
Text Solution
Verified by ExpertsThe correct answer is:
D
\(\theta = \tan^{-1} \left( \frac{v^2}{rg} \right) = \tan^{-1} \left[ \frac{(14 \sqrt{3})^2}{20 \sqrt{3} \times 9.8} \right] = \tan^{-1} [\sqrt{3}]\) = 60^\circ
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