Q. A motor cyclist moving with a velocity of $72\ km/hour$ on a flat road takes a turn on the road at a point where the radius of curvature of the road is $20\ m$. The acceleration due to gravity is $10\ m/\sec^{2}\ $. In order to avoid skidding, he must not bend with respect to the vertical plane by an angle greater than
A) $\theta = \tan^{- 1}6$
B) $\theta = \tan^{- 1}2$
C) $\theta = \tan^{- 1}{25.92}$
D) $\theta = \tan^{- 1}4$
Explanation
(b) $v = 72km/hour = 20m/sec$
$\theta = \tan^{- 1}\left( \frac{v^{2}}{rg} \right) = \tan^{- 1}\left( \frac{20 \times 20}{20 \times 20} \right) = \tan^{- 1}{(2)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.