A train is moving with a speed of 12 mathrm{~m} / mathrm{s} on rails which are 1.5 mathrm{~m} apart.

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4-1.Newton's Laws of Motion

hard

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

A $6.0 \mathrm{~cm}$

B $5.4 \mathrm{~cm}$

C$4.8 \mathrm{~cm}$

D $4.2 \mathrm{~cm}$

(JEE MAIN-2024)

Solution

$\tan \theta=\frac{\mathrm{v}^2}{\mathrm{Rg}}=\frac{12 \times 12}{10 \times 400}$
$ \tan \theta=\frac{\mathrm{h}}{1.5} $
$ \Rightarrow \frac{\mathrm{h}}{1.5}=\frac{144}{4000} $
$ \mathrm{~h}=5.4 \mathrm{~cm}$

Standard 11

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

Solution

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