What is the angle of banking of a railway track of radius of curvature 250m, if the maximum velocity of the train is 90 km / hr .? ( use` g= 10 m/s^(2))`

Options

A. `theta = tan ^(-1) ((1)/(2))`
B. `theta = tan ^(-1) ((1)/(3))`
C. `theta = tan ^(-1) ((1)/(4))`
D. ` theta =tan ^(-1)((1)/(5))`

Explanation

To find the angle of banking of a railway track, we can follow these steps:

### Step 1: Convert the velocity from km/hr to m/s
The maximum velocity of the train is given as 90 km/hr. To convert this to meters per second, we use the conversion factor:
\[ \text{Velocity in m/s} = \text{Velocity in km/hr} \times \frac{5}{18} \]
Calculating this:
\[ V = 90 \times \frac{5}{18} = 25 \text{ m/s} \]

### Step 2: Identify the radius of curvature and gravitational acceleration
The radius of curvature \( R \) is given as 250 m, and the acceleration due to gravity \( g \) is given as \( 10 \, \text{m/s}^2 \).

### Step 3: Use the formula for the angle of banking
The angle of banking \( \theta \) can be found using the formula:
\[ \tan \theta = \frac{V^2}{Rg} \]
Substituting the known values:
\[ \tan \theta = \frac{(25)^2}{250 \times 10} \]

### Step 4: Calculate \( V^2 \) and \( Rg \)
Calculating \( V^2 \):
\[ V^2 = 25^2 = 625 \]
Calculating \( Rg \):
\[ Rg = 250 \times 10 = 2500 \]

### Step 5: Substitute and simplify
Now substituting these values into the equation:
\[ \tan \theta = \frac{625}{2500} \]
Simplifying this:
\[ \tan \theta = \frac{1}{4} \]

### Step 6: Find the angle \( \theta \)
To find \( \theta \), we take the arctangent:
\[ \theta = \tan^{-1}\left(\frac{1}{4}\right) \]

### Final Result
Thus, the angle of banking \( \theta \) is:
\[ \theta = \tan^{-1}\left(\frac{1}{4}\right) \]