AP EAMCET · PHYSICS · Laws of Motion
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 \(90 \mathrm{~km} / \mathrm{h}\) ? \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
- A \(\theta=\tan ^{-1}\left(\frac{1}{2}\right)\)
- B \(\theta=\tan ^{-1}\left(\frac{1}{3}\right)\)
- C \(\theta=\tan ^{-1}\left(\frac{1}{4}\right)\)
- D \(\theta=\tan ^{-1}\left(\frac{1}{5}\right)\)
Answer & Solution
Correct Answer
(C) \(\theta=\tan ^{-1}\left(\frac{1}{4}\right)\)
Step-by-step Solution
Detailed explanation
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…
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