Q.Under what condition will a car skid on a leveled circular road?

On a level (unbanked) circular road, the entire centripetal force needed for a vehicle to turn safely, $F_{cp}=mv^2/r$, must come from static friction between the tyres and the road surface, which can supply at most $f_{s,max}=\mu_smg$ (since the normal force simply equals $mg$ on a flat road).

This gives a precise, calculable safe-turning condition: as long as $\frac{mv^2}{r}\le\mu_smg\quad\text{i.e.}\quad v\le\sqrt{\mu_srg},$ friction can supply the required centripetal force and the vehicle turns without incident. The moment the vehicle's speed exceeds $\sqrt{\mu_srg}$, friction can no longer keep pace with the demand for centripetal force, and the vehicle skids — its actual path curves less sharply than the road itself, carrying it outward off the intended track. …