The distance between two rails of railway track is $1.6m$ along a curve of radius $800m$. The outer rail is raised above the inner rail by $10cm$. What will be the maximum speed with which a train can be safely driven along the curve?
$\begin{align}
& A.22.2m{{s}^{-1}} \\
& B.12.2m{{s}^{-1}} \\
& C.42.2m{{s}^{-1}} \\
& D.90.9m{{s}^{-1}} \\
\end{align}$
Answer
615.3k+ views
Hint: First of all find out the banking angle. This can be found by
Dividing the distance with which the outer rail is raised by the distance between the two rails.
Then substitute the banking angle value in the standard formula which is given as,
$\tan \phi =\dfrac{{{V}^{2}}}{rg}$
Where $V$ be the maximum speed of the train, $g$ be the acceleration due to gravity and $r$be the radius of the curve formed.
Substituting all the values in it and rearranging the equation will give the correct answer.
Complete step by step answer:
First of all let us mention what all are given in the question,
The distance between the two railway tracks is
$d=1.6m$
The radius of the curve can be written as,
$r=800m$
The outer rail is raised above the inner rail by a value of,
$x=10cm$
First of all let us find out the banking angle. It is obtained by dividing the distance with which the outer rail is raised by the distance between the two rails.
Therefore we can write that,
$\tan \phi =\dfrac{x}{d}$
Substituting the values in it,
$\tan \phi =\dfrac{0.1}{1.6}=\dfrac{1}{16}$
Now let us apply this in the standard equation of banking angle which can be written as,
$\tan \phi =\dfrac{{{V}^{2}}}{rg}$
Substituting the values in it,
$\dfrac{1}{16}=\dfrac{{{V}^{2}}}{800\times 10}$
Rearranging the equation in terms of maximum speed will be,
$\dfrac{800\times 10}{16}={{V}^{2}}$
Simplifying this equation will give,
${{V}^{2}}=500$
Taking the square root of this will give the answer,
$V=\sqrt{500}=10\sqrt{5}m{{s}^{-1}}$
$V=22.2m{{s}^{-1}}$
So, the correct answer is “Option A”.
Note: The banking angle is defined as the angle at which a vehicle is being inclined about its longitudinal axis in accordance to the plane of the curved path. When the force of friction is not strong, then the vehicle will surely skid. This phenomenon is known as banking of roads.
Dividing the distance with which the outer rail is raised by the distance between the two rails.
Then substitute the banking angle value in the standard formula which is given as,
$\tan \phi =\dfrac{{{V}^{2}}}{rg}$
Where $V$ be the maximum speed of the train, $g$ be the acceleration due to gravity and $r$be the radius of the curve formed.
Substituting all the values in it and rearranging the equation will give the correct answer.
Complete step by step answer:
First of all let us mention what all are given in the question,
The distance between the two railway tracks is
$d=1.6m$
The radius of the curve can be written as,
$r=800m$
The outer rail is raised above the inner rail by a value of,
$x=10cm$
First of all let us find out the banking angle. It is obtained by dividing the distance with which the outer rail is raised by the distance between the two rails.
Therefore we can write that,
$\tan \phi =\dfrac{x}{d}$
Substituting the values in it,
$\tan \phi =\dfrac{0.1}{1.6}=\dfrac{1}{16}$
Now let us apply this in the standard equation of banking angle which can be written as,
$\tan \phi =\dfrac{{{V}^{2}}}{rg}$
Substituting the values in it,
$\dfrac{1}{16}=\dfrac{{{V}^{2}}}{800\times 10}$
Rearranging the equation in terms of maximum speed will be,
$\dfrac{800\times 10}{16}={{V}^{2}}$
Simplifying this equation will give,
${{V}^{2}}=500$
Taking the square root of this will give the answer,
$V=\sqrt{500}=10\sqrt{5}m{{s}^{-1}}$
$V=22.2m{{s}^{-1}}$
So, the correct answer is “Option A”.
Note: The banking angle is defined as the angle at which a vehicle is being inclined about its longitudinal axis in accordance to the plane of the curved path. When the force of friction is not strong, then the vehicle will surely skid. This phenomenon is known as banking of roads.
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