When the road is dry and the coefficient of friction is μ, the maximum speed of a car in a circular path is 10 m /s . If the road becomes wet and μ' = μ/2 , what is the maximum speed permitted

The correct option is 5√2 m / s.

To find the maximum speed permitted on the wet road, let's look at the physics of a car moving in a circular path on a horizontal road.

When a car of mass m travels along a circular path of radius R, the necessary centripetal force is provided by the static friction between the car's tires and the road surface. The maximum force of static friction is given by:

$f_{\max }=\mu N=\mu mg$
where:
• μ is the coefficient of static friction,
• N = m g is the normal reaction force, and
• g is the acceleration due to gravity.

The centripetal force required to keep the car moving in a circle of radius R at a speed v is:

$F_c=\frac{m{v}^2}{R}$
For the car to navigate the turn safely without slipping, the centripetal force must not exceed the maximum friction force:

$\frac{m{v}^2}{R}\le \mu mg$
Simplifying this inequality for the maximum speed ($v_{\max }$):

$v_{\max }^2=\mu gR$
$v_{\max }=\sqrt{\mu gR}$
From this relation, we can see that the maximum speed is directly proportional to the square root of the coefficient of friction ($v_{\max }\propto \sqrt{\mu }$).

Now, let us compare the two cases:

Case 1: Dry Road
The coefficient of friction is μ and the maximum speed $v_1=10\text{m/s}$:

$10=\sqrt{\mu gR}$       --- (Equation 1)

Case 2: Wet Road
The coefficient of friction becomes ${\mu }^{\prime }=\frac{\mu }{2}$. Let the new maximum speed be $v_2$:

$v_2=\sqrt{{\mu }^{\prime }gR}=\sqrt{\frac{\mu }{2}gR}$       --- (Equation 2)

Dividing Equation 2 by Equation 1 to find the relationship between the speeds:

$\frac{v_2}{10}=\frac{\sqrt{\frac{\mu }{2}gR}}{\sqrt{\mu gR}}$
$\frac{v_2}{10}=\frac{1}{\sqrt{2}}$
$v_2=\frac{10}{\sqrt{2}}$
Rationalizing the denominator:

$v_2=\frac{10\times \sqrt{2}}{2}=5\sqrt{2}\text{m/s}$
Thus, the maximum speed permitted on the wet road is 5√2 m / s.