Q. Keeping the angle of banking of the road constant, the maximum speed of the vehicles is to be increased by $10\%$ . The radius of curvature of the road will have to be changed from $20 \, m$ to
NTA AbhyasNTA Abhyas 2022
Solution:
The banking angle $\theta $ is given by,
$tan\theta =\frac{v^{2}}{r g}$ ; where $v$ is velocity, $r$ is radius and $g$ is acceleration due to gravity.
$rgtan\theta =v^{2}$
$\Rightarrow r \propto v^{2} \, \, $
Hence, the ratio of the radius will be,
$\Rightarrow \frac{r_{2}}{r_{1}}=\left(\frac{v_{2}}{v_{1}}\right)^{2}$
$\Rightarrow \frac{r_{2}}{20}=\left(\frac{\left(v + \frac{v}{10}\right)}{v}\right)^{2}$
$\Rightarrow r_{2}=24.2 \, m$ .
$tan\theta =\frac{v^{2}}{r g}$ ; where $v$ is velocity, $r$ is radius and $g$ is acceleration due to gravity.
$rgtan\theta =v^{2}$
$\Rightarrow r \propto v^{2} \, \, $
Hence, the ratio of the radius will be,
$\Rightarrow \frac{r_{2}}{r_{1}}=\left(\frac{v_{2}}{v_{1}}\right)^{2}$
$\Rightarrow \frac{r_{2}}{20}=\left(\frac{\left(v + \frac{v}{10}\right)}{v}\right)^{2}$
$\Rightarrow r_{2}=24.2 \, m$ .