The coefficient of static friction between the road and tyres of a car is 0.4 . The maximum permissible…
The coefficient of static friction between the road and tyres of a car is 0.4 . The maximum permissible speed of the car is $10 \mathrm{~ms}^{-1}$ on curved unbanked road. Then the maximum radius of curvature of the road is
(acceleration due to gravity $=10 \mathrm{~ms}^{-2}$ )
$10 \sqrt{5} \mathrm{~m}$
$25 \mathrm{~m}$
$20 \sqrt{2} \mathrm{~m}$
$30 \mathrm{~m}$
Solution
The necessary centripetal force is provided by friction force
$\begin{aligned} & \text { So, } \frac{\mathrm{mV}^2}{\mathrm{r}}=\mu \mathrm{mg} \\ & \Rightarrow \mathrm{V}=\sqrt{\mu \mathrm{rg}} \\ & \Rightarrow 10=\sqrt{0.4 \times \mathrm{r} \times 10} \\ & \Rightarrow 100=4 \mathrm{r} \\ & \Rightarrow \mathrm{r}=25 \mathrm{~m}\end{aligned}$