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Physics · Horizontal Circular Motion

JEE Main 2024 — 27 January, Shift 1 — Question 34

A train is moving with a speed of 12 m/s12 \mathrm{~m} / \mathrm{s} on rails which are 1.5 m apart. To negotiate a curve radius 400 m , the height by which the outer rail should be raised with respect to the inner rail is (Given, g=\mathrm{g}= 10 m/s2):\left.10 \mathrm{~m} / \mathrm{s}^{2}\right):

  1. Option A:

    6.0 cm

  2. Option B:

    5.4 cm

    Correct
  3. Option C:

    4.8 cm

  4. Option D:

    4.2 cm

Answer: B

Step-by-step solution

tan⁡θ=v2Rg=12×1210×400\tan \theta=\frac{\mathrm{v}^{2}}{\mathrm{Rg}}=\frac{12 \times 12}{10 \times 400}

tan⁡θ=h1.5\tan \theta=\frac{\mathrm{h}}{1.5}

⇒h1.5=1444000\Rightarrow \frac{\mathrm{h}}{1.5}=\frac{144}{4000}

h=5.4 cm\mathrm{h}=5.4 \mathrm{~cm}"

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
27 January, Shift 1
Subject
Physics
Chapter
Horizontal Circular Motion
Topic
Banking of Roads
A train is moving with a speed of 12 m / s on rails which are 1.5 m… | JEE Main 2024 PYQ with Solution · DhiX AI