Track Geometry, Creep and Cant

JKSSB Junior Engineer (Civil) · Railway Engineering · 19 practice questions with answers and explanations: 9 Easy, 9 Medium, 1 Hard.

Sample questions

Sample Easy question 1

Cant on a railway curve is also called:

  1. Superelevation
  2. Coning
  3. Camber
  4. Gradient
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Answer: A. It is the raising of the outer rail above the inner rail.

Sample Easy question 2

The formula for the cant on a railway curve is:

  1. C = G V ÷ 127 R
  2. C = G R ÷ 127 V²
  3. C = 127 V² ÷ G R
  4. C = G V² ÷ 127 R
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Answer: D. G is in mm, V in km/h and R in metres.

Sample Easy question 3

The maximum cant allowed on the broad gauge is:

  1. 50 mm
  2. 100 mm
  3. 200 mm
  4. 165 mm
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Answer: D. A sharper curve needs a lower speed, because the cant cannot exceed this limit.

Sample Medium question 4

For G = 1750 mm, V = 60 km/h and R = 400 m, the equilibrium cant is about:

  1. 62 mm
  2. 248 mm
  3. 124 mm
  4. 186 mm
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Answer: C. C = G V² ÷ 127 R = 1750 × 60² ÷ (127 × 400) = 124 mm.

Sample Medium question 5

The equilibrium cant needed for a speed is 110 mm and the cant provided is 80 mm. The cant deficiency is:

  1. 190 mm
  2. 110 mm
  3. 30 mm
  4. 80 mm
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Answer: C. Cant deficiency = equilibrium cant − actual cant, and it must not exceed 75 mm on the broad gauge.

Sample Medium question 6

The length of the transition curve by L = 0.008 Ca Vm, for a cant of 100 mm and a maximum speed of 80 km/h, is:

  1. 100 m
  2. 128 m
  3. 64 m
  4. 32 m
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Answer: C. L = 0.008 × 100 × 80 = 64 m.

Sample Hard question 7

For G = 1750 mm, V = 100 km/h and a cant of 140 mm, the radius of the curve R = G V² ÷ 127 C is about:

  1. 984 m
  2. 492 m
  3. 9.8 m
  4. 1,969 m
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Answer: A. R = 1750 × 100² ÷ (127 × 140) = 984 m.

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