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)]`

Answer

A

Explanation

Earth rotates from west to east. Net velocity of train `= (omega R - theta)^(2)`
Now, `mg - N = (m(omega R - theta)^(2))/(R )`
`:. N = mg - (m(omega R - nu)^(2))/(R )`
`= mg[1- (omega^(2)R^(2) + nu^(2) - 2 nu R omega)/(gR)]`
`= mg[1- (omega (omega R - 2nu))/(g) - nu^(2)/(Rg)]`