A car can safely negotiate a flat circular turn at a maximum speed...

The maximum safe speed of a car on a flat circular track is given by $v = \sqrt{\mu_s r g}$, where $\mu_s$ is the coefficient of static friction, $r$ is the radius of the turn, and $g$ is the acceleration due to gravity.

From the formula, the maximum safe speed is directly proportional to the square root of the coefficient of static friction:

$v \propto \sqrt{\mu_s}$

Let the initial maximum speed be $v_1 = 90\text{ km/h}$ and the initial coefficient of friction be $\mu_1$.

The new coefficient of friction is $\mu_2 = \dfrac{1}{9}\mu_1$.

The new maximum safe speed $v_2$ is:

$\dfrac{v_2}{v_1} = \sqrt{\dfrac{\mu_2}{\mu_1}}$

$\dfrac{v_2}{90} = \sqrt{\dfrac{1}{9}}$

$v_2 = 90 \times \dfrac{1}{3} = 30\text{ km/h}$

Answer: $30\text{ km/h}$