Question

When the road is dry and coefficient of static friction is $\mu$, the maximum speed of a car in the circular path is $10m{s}^{-1}$. If the road becomes wet and ${\mu }^{\prime }=\frac{\mu }{2}$, what is the maximum permitted speed for the car?

• A. $5m{s}^{-1}$
• B. $10m{s}^{-1}$
• C. $10\sqrt{2}m{s}^{-1}$
• D. $5\sqrt{2}m{s}^{-1}$

Answer 1

The correct option is D $5\sqrt{2}m{s}^{-1}$

To avoid skidding while the road is wet, static friction will act at maximum value to provide the necessary centripetal force for circular path.


So, we have $f=\mu N=\frac{m{v}_{max}^2}{R}$
or, $\mu mg=\frac{m{v}_{max}^2}{R}$
or, the maximum speed without skidding is
$v_{max}=\sqrt{\mu Rg}$
$\therefore \frac{v_{2max}}{v_{1max}}=\sqrt{\frac{{\mu }_2}{{\mu }_1}}=\sqrt{\frac{\mu }{2\mu }}=\frac{1}{\sqrt{2}}$
$\Rightarrow v_{2max}=\frac{v_{1max}}{\sqrt{2}}=\frac{10}{\sqrt{2}}=5\sqrt{2}m{s}^{-1}$