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JEE Mains · Physics · STD 11 - Friction
  • JEE Mains 2024
  • Hard
  • MCQ

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, \(g=\) \(10 \mathrm{~m} / \mathrm{s}^2\) ) :

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Answer & Solution

Correct Answer

(B)  \(5.4 \mathrm{~cm}\)

Step-by-step Solution

Detailed explanation

\(\tan \theta=\frac{\mathrm{v}^2}{\mathrm{Rg}}=\frac{12 \times 12}{10 \times 400}\)

\( \tan \theta=\frac{\mathrm{h}}{1.5} \)

\( \Rightarrow \frac{\mathrm{h}}{1.5}=\frac{144}{4000} \)

\( \mathrm{~h}=5.4 \mathrm{~cm}\)

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