JEE Mains · Physics · STD 11 - 4.2 friction
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, \(g=\) \(10 \mathrm{~m} / \mathrm{s}^2\) ) :
- A \(6.0 \mathrm{~cm}\)
- B \(5.4 \mathrm{~cm}\)
- C \(4.8 \mathrm{~cm}\)
- D \(4.2 \mathrm{~cm}\)
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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