question_answer

A road is 8 m wide. Its radius of curvature is 40 m. The outer edge is above the lower edge by a distance of 1.2 m. This road is most suited for a velocity of

A) 5.7 m/s                                

B) 7.7 m/s

C) 36.1 m/s                              

D) 9.7 m/s

Correct Answer: B

Solution :

\[\tan \theta =\frac{{{v}^{2}}}{rg}\] \[\sin \theta =\frac{{{v}^{2}}}{rg}\] \[(\sin \theta \approx \tan \theta ,for\,small\,\theta )\] \[\therefore \] \[\frac{1.2}{8}=\frac{{{v}^{2}}}{40\times 9.8}\] \[{{v}^{2}}=1.2\times 49\] \[v=\sqrt{1.2\times 49}\] \[=\sqrt{58.8}=7.7\,m/s\]