Question

A motor cyclist moving with a velocity of $72\text{km/hr}$ on flat road takes a turn on the road at a point where the radius of curvature of the road is $20\text{m}$. The acceleration due to gravity is $10{\text{m/s}}^2$. 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. $\theta ={\tan }^{-1}4$

Answer 1

The correct option is B $\theta ={\tan }^{-1}2$

Given, speed of motor cyclist $\left(v\right)=72\text{km/hr}=72\times \frac{5}{18}=20\text{m/s}$

Radius of curvature $\left(r\right)=20\text{m}$
Angle made by motor cyclist with vertical
${\theta }_{\text{min}}={\tan }^{-1}\left(\frac{v^2}{rg}\right)$
$={\tan }^{-1}\left(\frac{(20{)}^2}{20\times 10}\right)={\tan }^{-1}\left(\frac{400}{200}\right)$
$={\tan }^{-1}\left(2\right)$

Hence option B is the correct answer