Question

A train has to negotiate a curve of radius $2000\text{m}$. By how much should be the outer rail be raised with respect to the inner rail for a safe travelling speed of $72{\text{kmh}}^{-1}$?
The distance between the rails is $1\text{m}$.
$($Take $g=10{\text{m/s}}^2)$

• A. $1\text{cm}$
• B. $3\text{cm}$
• C. $2\text{cm}$
• D. $4.5\text{m}$

Answer 1

The correct option is C $2\text{cm}$


Given,
$v=72{\text{km h}}^{-1}=72\times \frac{5}{18}=20{\text{m s}}^{-1}$
Radius of curve, $R=2000\text{m};l=1\text{m}$

Let $h$ be the required height.

Applying Newton's $2^{nd}$ law on train,

$N\cos \theta =mg$

$N\sin \theta =\frac{m{v}^2}{R}$

From both equation, we get after diving

$\tan \theta =\frac{v^2}{Rg}.........\left(1\right)$

Also from diagram,

$\tan \theta =\frac{h}{l}.......\left(2\right)$

From eqs. $\left(1\right)\&\left(2\right)$,

$\frac{h}{l}=\frac{v^2}{gR}$

$h=\frac{v^{2}l}{gR}$

After putting the values, we get

$h=\frac{(20{)}^2\times 1}{10\times 2000}=\frac{400}{20000}$

$\therefore h=\frac{2}{100}\text{m}=2\text{cm}$

Hence, option $\left(c\right)$ is correct.