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A train of mass $m$ moves with a velocity $v$ on the equator from east to west. If $ω$ 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

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$m g [1 - 2 \frac{(ω R - v) ω}{g} - \frac{v^{2}}{R g}]$

$m g [1 - \frac{(ω R + 2 v) ω}{g} - \frac{v^{2}}{R g}]$

$m g [1 - \frac{(ω R - 2 v) ω}{g} - \frac{v^{2}}{R g}]$

answer is A.

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Detailed Solution

Earth rotates from west to east.

Net velocity of train $= ω R - v$

Now, $m g - N = \frac{m (ω R - v)^{2}}{R}$

$∴ N = m g - \frac{m (ω R - v)^{2}}{R}$

$= m g [1 - \frac{ω^{2} R^{2} + v^{2} - 2 v R ω}{g R}]$

$= m g [1 - \frac{ω (ω R - 2 v)}{g} - \frac{v^{3}}{R g}]$

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