Geometric Design

JKSSB Junior Engineer (Civil) · Highway Engineering · 14 practice questions with answers and explanations: 6 Easy, 6 Medium, 2 Hard.

Sample questions

Sample Easy question 1

The ideal alignment of a highway should be:

  1. Short, easy, safe and economical
  2. As long as possible
  3. Always straight regardless of cost
  4. Along existing footpaths only
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Answer: A. These are the four requirements of a good alignment.

Sample Easy question 2

The maximum superelevation recommended by the IRC for plain and rolling terrain is:

  1. 10% for all terrain
  2. 7%
  3. 15%
  4. 3%
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Answer: B. On hill roads not bound by snow it is 10%.

Sample Easy question 3

A valley curve on a highway is designed mainly on the basis of:

  1. Headlight sight distance at night
  2. Daylight stopping sight distance
  3. The width of the road
  4. The type of pavement
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Answer: A. A summit curve is designed for stopping or overtaking sight distance.

Sample Medium question 4

Mechanical widening of a pavement on a curve of radius 100 m for 2 lanes with a wheel base of 6 m is about:

  1. 0.18 m
  2. 0.72 m
  3. 0.36 m
  4. 2.27 m
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Answer: C. Wm = n l² ÷ 2R = 2 × 6² ÷ (2 × 100) = 0.36 m.

Sample Medium question 5

Psychological widening on a curve of radius 100 m for a design speed of 60 km/h is about (Wps = V ÷ 2.64√R):

  1. 0.36 m
  2. 2.27 m
  3. 4.55 m
  4. 1.14 m
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Answer: B. Wps = 60 ÷ (2.64 × √100) = 2.27 m.

Sample Medium question 6

A gradient of 1 in 20 is equal to:

  1. 20%
  2. 5%
  3. 2%
  4. 0.5%
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Answer: B. 100 ÷ 20 = 5%.

Sample Hard question 7

On a curve of radius 150 m with a design speed of 70 km/h, for 2 lanes and a wheel base of 6.1 m, the total extra widening (mechanical plus psychological) is about:

  1. 0.2 m
  2. 2.4 m
  3. 2.2 m
  4. 4.8 m
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Answer: B. Wm = n l² ÷ 2R = 0.25 m, Wps = V ÷ 2.64√R = 2.16 m, so the total is 2.41 m.

Sample Hard question 8

With a rate of change of centrifugal acceleration C = 80 ÷ (75 + V) = 0.593, the length of the transition curve for V = 60 km/h and R = 300 m (L = 0.0215 V³ ÷ C R) is about:

  1. 52 m
  2. 60 m
  3. 26 m
  4. 13 m
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Answer: C. L = 0.0215 × 60³ ÷ (0.593 × 300) = 26.1 m.

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