A turn has a radius of `10m` if a vehicle goes round it at an average speed of `18km//h` , what should be the proper angle of banking?

Options

A. `45^(@)`
B. `tan^(-1) (1)/(4)`
C. `30^(@)`
D. `tan^(-1) (1)/(2)`

Explanation

To find the proper angle of banking for a vehicle going around a turn, we can follow these steps:

### Step 1: Convert the speed from km/h to m/s
The speed of the vehicle is given as 18 km/h. To convert this to meters per second, we use the conversion factor:
\[ \text{Speed in m/s} = \text{Speed in km/h} \times \frac{5}{18} \]
Calculating this gives:
\[ \text{Speed} = 18 \times \frac{5}{18} = 5 \text{ m/s} \]

### Step 2: Identify the radius of the turn
The radius of the turn is given as \( r = 10 \, m \).

### Step 3: Set up the equations for forces
When a vehicle is banking on a turn, the forces acting on it can be resolved into two components:
1. The vertical component of the normal force balances the weight of the vehicle.
2. The horizontal component of the normal force provides the necessary centripetal force for circular motion.

Let \( N \) be the normal force and \( \theta \) be the banking angle. The equations can be written as:
- Vertical component: \( N \cos \theta = mg \)
- Horizontal component (centripetal force): \( N \sin \theta = \frac{mv^2}{r} \)

### Step 4: Divide the equations
Dividing the two equations gives:
\[ \frac{N \sin \theta}{N \cos \theta} = \frac{\frac{mv^2}{r}}{mg} \]
This simplifies to:
\[ \tan \theta = \frac{v^2}{rg} \]

### Step 5: Substitute the known values
Now, we can substitute the values into the equation:
- \( v = 5 \, m/s \)
- \( r = 10 \, m \)
- \( g = 10 \, m/s^2 \) (approximate value for acceleration due to gravity)

Substituting these values gives:
\[ \tan \theta = \frac{5^2}{10 \times 10} = \frac{25}{100} = \frac{1}{4} \]

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

### Conclusion
The proper angle of banking for the vehicle is:
\[ \theta = \tan^{-1}\left(\frac{1}{4}\right) \]