A car sometimes overturns while taking a turn. When it overturns, it is

The correct option is: The inner wheel which leaves the ground first

To understand why the inner wheel of a car leaves the ground first when it overturns while taking a turn, let us analyze the forces and torques acting on the car.

Consider a car of mass $m$ moving with a constant speed $v$ along a circular path of radius $r$. Let the distance between the inner and outer wheels (track width) be $2a$ and the height of the center of gravity (CG) of the car above the road surface be $h$.

When the car takes a turn, it experiences a centrifugal force acting horizontally outwards through its center of gravity. The magnitude of this centrifugal force is:
$F=\frac{m{v}^2}{r}$
This force creates an overturning torque (moment of force) about the point of contact of the wheels, trying to tip the car outwards.

Let $R_1$ be the normal reaction on the inner wheels and $R_2$ be the normal reaction on the outer wheels. For vertical equilibrium of the car, the sum of the normal reactions must balance the weight of the car:
$R_1+R_2=mg$

Taking moments about the center of gravity (CG) for rotational equilibrium, the overturning torque must be balanced by the restoring torque created by the normal reactions:
$(R_2-R_1)a=F·h$
Substituting the expression for the centrifugal force $F$:
$R_2-R_1=\frac{m{v}^{2}h}{ar}$

By solving these two equations for $R_1$ and $R_2$, we obtain:
$R_1=\frac{mg}{2}-\frac{m{v}^{2}h}{2ar}$
and
$R_2=\frac{mg}{2}+\frac{m{v}^{2}h}{2ar}$

From the expression for $R_1$, we can see that as the speed $v$ of the car increases, the normal reaction on the inner wheels decreases. When the speed reaches a critical value where the car is just about to overturn, the normal reaction on the inner wheels becomes zero ($R_1=0$). This indicates that the inner wheel loses contact with the ground first.