A car is moving on a circular path and takes a turn. If R1 and R2 be the reactions on the inner and outer wheels respectively, then

To understand the reactions on the inner and outer wheels of a car taking a turn on a circular path, we need to analyze the forces acting on the car. Let us denote the reactions on the inner and outer wheels as $R_1$ and $R_2$ respectively.

When a car of mass $m$ moves on a horizontal circular path of radius $r$ with velocity $v$, it experiences a centripetal force. This force is provided by the friction between the tyres and the road, acting towards the center of the circular path. The frictional force $F$ can be written as:
$F=\frac{m{v}^2}{r}$

Let $h$ be the height of the center of gravity of the car above the road surface, and $2a$ be the distance between the inner and outer wheels (track width). The center of gravity is at a distance $a$ from both the inner and outer wheels.

Considering the vertical equilibrium of the car, the total normal reaction must balance the weight of the car:
$R_1+R_2=mg$

Now, let us take the torque (moment of forces) about the center of gravity of the car. For rotational equilibrium, the net torque must be zero. The forces creating torque about the center of gravity are:
1. The normal reaction of the inner wheel $R_1$, which acts upwards at a distance $a$, producing a clockwise torque: $R_1\cdot a$.
2. The normal reaction of the outer wheel $R_2$, which acts upwards at a distance $a$, producing a counter-clockwise torque: $R_2\cdot a$.
3. The centrifugal force (or the effect of the centripetal acceleration in the car's frame) $\frac{m{v}^2}{r}$ acting outwards through the center of gravity, which is balanced by the inward frictional force $F$ acting at the road level (distance $h$ below the center of gravity). This pair of forces creates an overturning couple (clockwise torque) equal to:
$F\cdot h=\frac{m{v}^{2}h}{r}$

Equating the clockwise and counter-clockwise torques about the center of gravity:
$R_2\cdot a=R_1\cdot a+\frac{m{v}^{2}h}{r}$
Dividing the entire equation by $a$ gives:
$R_2-R_1=\frac{m{v}^{2}h}{ra}$

Solving the system of equations for $R_1$ and $R_2$:
$R_1=\frac{mg}{2}-\frac{m{v}^{2}h}{2ra}$
$R_2=\frac{mg}{2}+\frac{m{v}^{2}h}{2ra}$

Since the term $\frac{m{v}^{2}h}{2ra}$ is strictly positive for a moving car ($v>0$), we can clearly see that:
$R_1<R_2$

Therefore, the reaction on the inner wheels $R_1$ is less than the reaction on the outer wheels $R_2$ when a car takes a turn on a circular path.