A cyclist is travelling with velocity v on a banked curved road of radius R. The angle through which the cyclist leans inwards is given by

Options

A. `tan theta =(Rg)/v^(2)`
B. `tan theta =v^(2) Rg`
C. `tan theta =(v^(2)g)/R`
D. `tan theta = v^(2)/(Rg)`

Explanation

To solve the problem of determining the angle through which a cyclist leans inward while traveling on a banked curved road, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Forces Acting on the Cyclist**:
- The forces acting on the cyclist are:
- The gravitational force \( mg \) acting downwards.
- The normal force \( N \) acting perpendicular to the surface of the banked road.

2. **Break Down the Forces**:
- The normal force \( N \) can be resolved into two components:
- A vertical component \( N \cos \theta \) that balances the weight of the cyclist.
- A horizontal component \( N \sin \theta \) that provides the necessary centripetal force for circular motion.

3. **Set Up the Equations**:
- For vertical forces (balancing the weight):
\[ N \cos \theta = mg \quad (1) \]
- For horizontal forces (providing centripetal acceleration):
\[ N \sin \theta = \frac{mv^2}{R} \quad (2) \]

4. **Divide the Equations**:
- Divide equation (2) by equation (1) to eliminate \( N \):
\[ \frac{N \sin \theta}{N \cos \theta} = \frac{\frac{mv^2}{R}}{mg} \]
- This simplifies to:
\[ \tan \theta = \frac{v^2}{gR} \quad (3) \]

5. **Solve for the Angle \( \theta \)**:
- Rearranging equation (3), we find:
\[ \theta = \tan^{-1}\left(\frac{v^2}{gR}\right) \]

### Final Answer:
The angle through which the cyclist leans inward is given by:
\[ \theta = \tan^{-1}\left(\frac{v^2}{gR}\right) \]