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Circular Motion question

2024 · 27 Jan · Shift 1 · Q66
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  5. /2024 · 27 Jan · Shift 1 · Q66

Circular Motion question

2024 · 27 Jan · Shift 1 · Q66

JEE MainPhysicsCircular MotionMCQ+4 / −1
A train is moving with a speed of 12 m/s12 \mathrm{~m} / \mathrm{s}12 m/s on rails which are 1.5 m1.5 \mathrm{~m}1.5 m apart. To negotiate a curve radius 400 m400 \mathrm{~m}400 m, the height by which the outer rail should be raised with respect to the inner rail is (Given, g=10 m/s2)g=10 \mathrm{~m} / \mathrm{s}^2)g=10 m/s2) :
  1. A
    6.0 cm
  2. B
    5.4 cm
  3. C
    4.8 cm
  4. D
    4.2 cm
View written solutionFree

Correct answer: B

  1. Condition for banking of railway track

For a train moving on a curved track, the required cant (height difference) hhh between the outer and inner rails is given by

anθ=v2rg an\theta = \frac{v^2}{rg}anθ=rgv2​

If the rails are separated by distance ddd, then for small angle θ\thetaθ,

tan⁡θ≈hd\tan\theta \approx \frac{h}{d}tanθ≈dh​

So,

hd=v2rg\frac{h}{d} = \frac{v^2}{rg}dh​=rgv2​

Hence,

h=d v2rgh = d\,\frac{v^2}{rg}h=drgv2​

  1. Substitute the given values

Given:

  • v=12 m/sv = 12\,\text{m/s}v=12m/s
  • d=1.5 md = 1.5\,\text{m}d=1.5m
  • r=400 mr = 400\,\text{m}r=400m
  • g=10 m/s2g = 10\,\text{m/s}^2g=10m/s2

Therefore,

h=1.5×122400×10h = 1.5\times \frac{12^2}{400\times 10}h=1.5×400×10122​

h=1.5×1444000h = 1.5\times \frac{144}{4000}h=1.5×4000144​

h=1.5×0.036h = 1.5\times 0.036h=1.5×0.036

h=0.054 mh = 0.054\,\text{m}h=0.054m

  1. Convert into cm

0.054 m=5.4 cm0.054\,\text{m} = 5.4\,\text{cm}0.054m=5.4cm

  1. Match with the options

Thus, the required height difference is

5.4 cm\boxed{5.4\,\text{cm}}5.4cm​

So the correct option is B.

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