Answer:

c. 0.816

Explanation:

Let the mass of car be ‘m’ and coefficient of static friction be ‘μ’.

Given:

Speed of the car (v) = 40.0 m/s

Radius of the curve (R) = 200 m

As the car is making a circular turn, the force acting on it is centripetal force which is given as:

Centripetal force is, [tex]F_c=\frac{mv^2}{R}[/tex]

The frictional force is given as:

Friction = Normal force × Coefficient of static friction

[tex]f=\mu N[/tex]

As there is no vertical motion, therefore, [tex]N=mg[/tex]. So,

[tex]f=\mu mg[/tex]

Now, the centripetal force is provided by the frictional force. Therefore,

Frictional force = Centripetal force

[tex]f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu=\frac{v^2}{Rg}[/tex]

Plug in the given values and solve for ‘μ’. This gives,

[tex]\mu = \frac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}\\\\\mu=\frac{1600\ m^2/s^2}{1960\ m^2/s^2}\\\\\mu=0.816[/tex]

Therefore, option (c) is correct.