A train is running at $20 \,m / s$ on a railway line with radius of curvature $40,000$ metres. The distance between the two rails is $1.5$ metres. For safe running of train the elevation of outer rail over the inner rail is ......$mm$ $\left( g =10 \,m / s ^2\right)$
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(c)
$\tan \theta=\frac{h}{d}=\frac{v^2}{r g}$
$h=\frac{(1.5)(20)(20)}{40,000 \times 10}$
$=1.5 \,mm$
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