I know that $N = W \cos \theta$ where $W = mg$, but while reading about circular motion of a car on the banked road, the two forces were balanced as $N \cos \theta = W$. How is this possible?
I know this has already been asked before, but I did not understand any of the answers.
You are confusing circular motion with motion on an inclined plane. The most important difference is where does the net force points to: (i) for the latter the net force is in the direction of the inclined plane, and (ii) for the former the net force is in the horizontal direction, i.e. towards the center of circular motion.
For the motion on an inclined plane, the object moves along the surface, i.e. there are forces in both vertical and horizontal directions whose resultant is in the direction of the inclined plane.
For the circular motion there is no net force in the vertical direction, i.e. there is only force component in the horizontal direction which is in the direction of the center of circular motion. Hence the term centripetal force or radial force.
The trick to understand this is that you start from the net (resultant) force and then solve what normal force must be in order to satisfy the initial assumption.
I will emphasize once more the initial assumption for circular motion on banked roads: (i) the car does not move in vertical direction, and (ii) there is net force in the direction of the center of circular motion.
Let’s say that the road is at angle $\theta$ to the horizontal surface. Then the road exerts normal force on the car at the angle $(90^\circ - \theta)$ to horizontal. Since the car is in equilibrium in the vertical direction, according to the first Newton’s law we have:
$$w - n \cdot \cos \theta = 0 \tag 1$$
where $w$ is the car’s weight:
$$w = m \cdot g \tag 2$$
The only force in the horizontal direction is the projection of the normal force to horizontal:
$$F_\text{net} = n \cdot \cos (90^\circ - \theta) = n \cdot \sin \theta \tag 3$$
The net force in horizontal direction is actually the centripetal force that acts always towards the center of circular motion. According to the second Newton’s law we have:
$$F_\text{net} = m \cdot a \tag 4$$
By combining all these equations we finally get the expression for the centripetal acceleration:
$$\boxed{a = g \cdot \tan \theta}$$
where the centripetal acceleration is defined as:
$$\boxed{a = \frac{v^2}{R}}$$
where $v$ is car's speed and $R$ is radius of the circular motion. Here you can find derivation of the expression for the centripetal acceleration: https://physics.stackexchange.com/a/685423/149541
but while reading about circular motion of a car on the banked road, the two forces were balanced as $N \cos \theta = W$. How is this possible?
It is one of the equations that are needed in order to determine the maximum speed the car can have without sliding on a frictionless banked turn.
The example you have been given is for a car on a banked, frictionless road. The free body diagram where friction is included is shown in the Figure below (source:http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/carbank.html).
If the road is frictionless, then $\mu_{s}=0$.
There is a net force (centripetal force) acting in the $x$-direction (along the radius of the turn towards the center) which is needed for the car to turn. For a frictionless surface
$$\sum F_{x}=m\frac{v^2}{r}=N\sin\theta$$ $$m\frac{v^2}{r}=N\sin\theta$$
For equilibrium in the $y$ direction, the net force acting in the $y$ direction is zero, or
$$\sum F_{y}=0=N\cos\theta - mg$$
And therefore,
$$N\cos\theta=mg$$
The maximum speed of the vehicle on a frictionless banked turn is obtained by solving the second and forth equations simultaneously (dividing the second by the fourth), which gives.
$$v_{max}=\sqrt {rg\tan\theta}$$
Hope this helps.
