Q. A road of 10 m width has radius of curvature 50 m. Its outer edge is raised above the inner edge by a distance of 1.5 m. The road is most suited for vehicles moving with velocity of
A) $8.5\ ms^{- 1}$
B) 6.5$\ ms^{- 1}$
C) 5.5$\ ms^{- 1}$
D) None of these
Explanation
(a) $\tan{\theta = \frac{h}{b} = \frac{v^{2}}{rg},}$
$v = \sqrt{\frac{hrh}{b}} = \sqrt{\frac{1.5 \times 50 \times 10}{10}} = 8.5\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.