What should be the elevation of outer track of the train to move in a circular path of radius R, width of the track is w (< < R) and speed of the train is v ? (Neglect friction)

The correct option is v²w / Rg.

To find the elevation of the outer track of the train, we analyze the banking of the circular track. When a train moves in a circular path of radius $R$ at a speed $v$, it requires a centripetal force to keep it in the circular path. In the absence of friction, this force is provided by the horizontal component of the normal reaction force from the rails, which is achieved by banking the track (raising the outer rail relative to the inner rail).

Let $\theta$ be the angle of banking of the track. The relationship between the angle of banking, speed of the train $v$, radius $R$, and acceleration due to gravity $g$ is given by:
$\tan \left(\theta \right)=\frac{v^2}{Rg}$

Let $h$ be the height (elevation) by which the outer rail is raised, and $w$ be the width of the track. From the geometry of the banked track, we can express the sine of the angle as:
$\sin \left(\theta \right)=\frac{h}{w}$

For a typical railway track, the elevation $h$ is much smaller than the width of the track $w$ ($h<<w$). This means that the angle of banking $\theta$ is very small. For very small angles, we can use the approximation:
$\tan \left(\theta \right)\approx \sin \left(\theta \right)$

Equating the two expressions, we get:
$\frac{h}{w}=\frac{v^2}{Rg}$

Solving for the elevation $h$ of the outer track gives:
$h=\frac{v^{2}w}{Rg}$