A train of mass `m` moves with a velocity `upsilon` on the equator from east to west. If `omega` is the angular speed of earth about its axis and `R` is the radius of the earth then the normal reaction acting on the train is

Options

A. `mg[1-((omega R - 2 upsilon) omega )/(g)-(upsilon^(2))/(Rg)]`
B. `mg[1- 2((omega R - upsilon) omega )/(g)-(upsilon^(2))/(Rg)]`
C. `mg[1- ((omega R - 2 upsilon) omega )/(g)-(upsilon^(2))/(Rg)]`
D. `mg[1-2 ((omega R - upsilon) omega )/(g)-(upsilon^(2))/(Rg)]`

Explanation

To solve the problem of finding the normal reaction acting on a train moving from east to west at the equator, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Forces Acting on the Train**:
- The gravitational force acting downward on the train is \( mg \) (where \( m \) is the mass of the train and \( g \) is the acceleration due to gravity).
- The normal reaction force \( N \) acts upward.

2. **Consider the Earth's Rotation**:
- The Earth rotates from west to east with an angular speed \( \omega \).
- The linear speed \( v \) of the train is directed from east to west.

3. **Calculate the Effective Velocity**:
- The effective velocity of the train relative to the Earth's surface can be expressed as:
\[ v_{\text{effective}} = \omega R - v \]
- Here, \( \omega R \) is the speed of a point on the equator due to Earth's rotation.

4. **Apply Newton's Second Law**:
- According to Newton's second law, the net force acting on the train in the vertical direction can be expressed as:
\[ mg - N = m \cdot a \]
- Where \( a \) is the centripetal acceleration due to the effective velocity:
\[ a = \frac{(v_{\text{effective}})^2}{R} = \frac{(\omega R - v)^2}{R} \]

5. **Substituting for Centripetal Acceleration**:
- Substitute \( a \) into the equation:
\[ mg - N = m \cdot \frac{(\omega R - v)^2}{R} \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ N = mg - m \cdot \frac{(\omega R - v)^2}{R} \]

7. **Simplifying the Expression**:
- Factor out \( m \):
\[ N = m \left( g - \frac{(\omega R - v)^2}{R} \right) \]
- Expanding \( (\omega R - v)^2 \):
\[ N = m \left( g - \frac{\omega^2 R^2 - 2\omega R v + v^2}{R} \right) \]
- This simplifies to:
\[ N = mg \left( 1 - \frac{\omega^2 R - 2v}{g} - \frac{v^2}{gR} \right) \]

8. **Final Expression for Normal Reaction**:
- Thus, the normal reaction \( N \) acting on the train is:
\[ N = mg \left( 1 - \frac{\omega^2 R - 2v}{g} - \frac{v^2}{gR} \right) \]

### Final Answer:
The normal reaction acting on the train is:
\[ N = mg \left( 1 - \frac{\omega^2 R - 2v}{g} - \frac{v^2}{gR} \right) \]