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What is the angle of banking of a railway track of radius of curvature $250 \mathrm{~m}$, if the maximum velocity of the train is $90 \mathrm{~km} / \mathrm{h}$ ? $\left(g=10 \mathrm{~ms}^{-2}\right)$
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Verified Answer
The correct answer is:
$\theta=\tan ^{-1}\left(\frac{1}{4}\right)$
Given, radius of curvature, $r=250 \mathrm{~m}$
Maximum velocity of train, $v=90 \mathrm{kmh}^{-1}$
$$
=\frac{90 \times 5}{18} \mathrm{~m} / \mathrm{s}=25 \mathrm{~m} / \mathrm{s}
$$
Let $\theta$ be the angle of banking of railway track.
Then, by using an expression of maximum safe speed,
$$
v_2=r g \tan \theta \Rightarrow \tan \theta=\frac{v^2}{r g}
$$
By substituting the values, we get
$$
\begin{aligned}
& \tan \theta=\frac{(25)^2}{250 \times 10}=\frac{625}{2500}=\frac{1}{4} \\
\therefore \quad \theta & =\tan ^{-1}\left(\frac{1}{4}\right)
\end{aligned}
$$
Maximum velocity of train, $v=90 \mathrm{kmh}^{-1}$
$$
=\frac{90 \times 5}{18} \mathrm{~m} / \mathrm{s}=25 \mathrm{~m} / \mathrm{s}
$$
Let $\theta$ be the angle of banking of railway track.
Then, by using an expression of maximum safe speed,
$$
v_2=r g \tan \theta \Rightarrow \tan \theta=\frac{v^2}{r g}
$$
By substituting the values, we get
$$
\begin{aligned}
& \tan \theta=\frac{(25)^2}{250 \times 10}=\frac{625}{2500}=\frac{1}{4} \\
\therefore \quad \theta & =\tan ^{-1}\left(\frac{1}{4}\right)
\end{aligned}
$$
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