Question

A car is moving on a levelled circular road of radius of curvature $300\text{m}$. If the coefficient of static friction is $0.3$ and acceleration due to gravity is $10m{s}^{-2}$, then maximum allowable speed for the car will be (in $km{h}^{-1})$

• A. $30$
• B. $81$
• C. $108$
• D. $162$

Answer 1

The correct option is C $108$

To avoid slipping, frictional force will act in a direction to provide the necessary centripetal force. i.e frictional force will act towards centre of circular curvature.
Here, $f$ is magnitude of frictional force.
$\therefore$ $f=\frac{m{v}_{max}^2}{r}$
$\therefore$ ${\mu }_{s}mg=\frac{m{v}_{max}^2}{r}$

Here, it is given that
$r=300\text{m},{\mu }_s=0.3,g=10m{s}^{-2}$
Applying the formula,
$v_{max}=\sqrt{{\mu }_{s}rg}=\sqrt{0.3\times 300\times 10}=30m{s}^{-1}$
Or, $30\times \frac{18}{5}km{h}^{-1}=108km{h}^{-1}$