A car of mass m attempts to navigate a circular road of radius r...

The road is banked at an angle $\theta$ such that $\tan \theta = \dfrac{v_0^2}{rg}$, where $v_0 = 40 \text{ km/hr}$.

When the car moves at a speed $v$, the forces acting on it are the normal force $N$ and gravity $mg$.

Resolving forces along the incline, the downward component of gravity is $mg \sin \theta$ and the required centripetal component is $\dfrac{mv^2}{r} \cos \theta$.

If $v < v_0$, $mg \sin \theta > \dfrac{mv^2}{r} \cos \theta$. Since friction is negligible, there is a net downward force along the incline, causing the car to slip downwards. Thus, option (1) is correct.

When the car moves at exactly $v = 40 \text{ km/hr}$, it does not slip. The normal force $N$ provides both the vertical balance and the centripetal force:

$N \cos \theta = mg$

$N \sin \theta = \dfrac{mv^2}{r}$

Squaring and adding these equations gives the total force exerted by the road on the car:

$N = \sqrt{(mg)^2 + \left(\dfrac{mv^2}{r}\right)^2}$

Since both $mg$ and $\dfrac{mv^2}{r}$ are non-zero positive quantities, $N > mg$ and $N > \dfrac{mv^2}{r}$. Thus, option (3) is correct.