Question 200 of 210

A train is moving with a speed of $12m{s}^{-1}$ on rails which are $1.5m$ apart. To negotiate a curve radius $400m$, the height by which the outer rail should be raised with respect to the inner rail is (Given, $g=$ $10m{s}^{-2}$):

A$6.0cm$B$5.4cm$C$4.8cm$D$4.2cm$

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Answer: (B) $5.4cm$

For the train, moving along the curved rails, it can be written, with reference to the figure below, that
$\tan \theta =\dfrac{{v}^{2}}{Rg}...\left(1\right)$

Also, from the above figure, it follows that
$\tan \theta =\dfrac{h}{1.5}...\left(2\right)$
Equations (1) and (2) imply that
$\dfrac{h}{1.5}=\dfrac{{v}^{2}}{Rg} \Rightarrow h=1.5\times \dfrac{{v}^{2}}{Rg} =1.5\times \dfrac{{\left(12m{s}^{-1}\right)}^{2}}{400m\times 10m{s}^{-2}} =0.054m =5.4cm$