The distance between two rails is 1.5m . The centre of gravity of the train at a height of 2m from the ground. The maximum speed of the train on a circular path of radius 120m can be

The correct option is 21 m /s.

Understanding the Physical Principles:
When a train moves along a circular path of radius $R$, it experiences a centrifugal force acting horizontally outwards through its center of gravity. This force tends to overturn the train about the outer rail. Simultaneously, the weight of the train acts vertically downwards through the center of gravity, creating a restoring torque about the outer rail that keeps the train stable.

To find the maximum safe speed before the train begins to overturn, we equate the overturning torque to the restoring torque about the outer rail:

$\text{Overturning Torque}=\text{Restoring Torque}$

$F_c·h=W·\frac{w}{2}$

Where:
- $F_c=\frac{m{v}^2}{R}$ is the centrifugal force,
- $h$ is the height of the center of gravity from the ground ($2\text{m}$),
- $W=mg$ is the weight of the train,
- $w$ is the distance between the two rails (gauge width = $1.5\text{m}$),
- $R$ is the radius of the circular path ($120\text{m}$),
- $g$ is the acceleration due to gravity (taken as $9.8{\text{m/s}}^2$).

Step-by-Step Derivation:
Substitute the forces into the torque equation:
$\left(\frac{m{v}^2}{R}\right)·h=mg·\frac{w}{2}$

Cancel the mass ($m$) from both sides and solve for velocity ($v$):
$v^2=\frac{g·R·w}{2·h}$

Substitute the given numerical values:
$v^2=\frac{9.8·120·1.5}{2·2}$

$v^2=\frac{9.8·180}{4}$

$v^2=9.8·45$

$v^2=441$

Taking the square root of both sides:
$v=\sqrt{441}=21\text{m/s}$

Therefore, the maximum speed of the train on the circular path to prevent it from overturning is 21 m/s.