A train is moving with a speed of 12 m/s on rails which are 1.5 m apart. To negotiate a curve of radius 400 m, the height by which the outer rail should be raised with respect to the inner rail is (Given, g = 10 m/s²): (a) 6.0 cm (b) 5.4 cm (c) 4.8 cm (d) 4.2 cm

Answer: (b). tan θ = v²/rg ≈ h/d ⇒ h = (d v²)/(rg) h = (1.5 × 144)/(400 × 10) = 0.054 m h = 5.4 cm So the answer is (b) 5.4 cm.

A train is moving with a speed of 12 m/s on rails which are 1.5 m… | JEE Main 2024 (27 Jan, Shift 1)

Q8JEE Main 2024 · 27 JAN · SHIFT 1Easy

Banking of Roads & Bridges

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

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Answer: (b)

$\tan\theta = \frac{v^2}{rg} \approx \frac{h}{d} \Rightarrow h = \frac{d v^2}{rg}$
$h = \frac{1.5 \times 144}{400 \times 10} = 0.054$ m
$h = 5.4$ cm
So the answer is (b) 5.4 cm.

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