Highway Cross-sectional Elements - IES Previous Year Questions

1. Consider the following types of roads in the same rainfall region:

  1. Water Bound Macadam roads
  2. Cement Concrete roads
  3. Bituminous high-speed roads
  4. Gravel roads

The correct sequence of the ascending order of steepness of camber of these roads is _____ .

[IES 1995]

  1. 4, 1, 3, 2
  2. 1, 4, 3, 2
  3. 1, 4, 2, 3
  4. 4, 1, 2, 3
Answer: Option A
Explanation:
  • For Cement Concrete roads, camber may be provided between 1 in 60 and 1 in 50.
  • For Bituminous roads, camber may be provided between 1 in 50 and 1 in 40.
  • For Water Bound Macadam roads, camber may be provided between 1 in 40 and 1 in 33.
  • For gravel roads, camber may be provided between 1 in 25 and 1 in 30.
Hence, the increasing order of steepness of camber for different roads is: Cement Concrete roads, Bituminous roads, Water Bound Macadam roads and Gravel roads.

2. A vehicle was stopped in two seconds by fully jamming the brakes. The skid marks measured 9.8 meters. The average skid resistance coefficient will be _____ .

[IES 1997]

  1. 0.7
  2. 0.5
  3. 0.4
  4. 0.25
Answer: Option B
Explanation:
The average skid resistance coefficient (f) can be calculated using the below formulae:
f =
a / g
, 
2L / gt2
, 
v2 / 2gL

Using the expression, f =
2L / gt2

⇒ f =
2 x 9.8 / 9.8 x 22

⇒ f = 0.5

3.  Brake is applied on a vehicle which then skids a distance of 16 m before coming to stop. If the developed average coefficient of friction between the tyres and the pavement is 0.4, then the speed of the vehicle before skidding have been nearly _____ .

[IES 1999]

  1. 20 kmph
  2. 30 kmph
  3. 40 kmph
  4. 50 kmph
Answer: Option C
Explanation:
The average skid resistance coefficient (f) can be calculated using the below formulae:
f =
a / g
, 
2L / gt2
, 
v2 / 2gL

Using the expression, f =
v2 / 2gL

⇒ 0.4 =
v2 / 2 x 10 x 16
(Use g = 10 m/s2)
⇒ v = 11.31 m/s (or) 40.73 kmph
⇒ v ≈ 40 kmph

4.  Assertion (A): Worn out (smooth) tyres offer higher friction factors on dry pavements than new tyres with treads well intact.

Reason (R): Reduced pneumatic pressure is held in tubes which carry smooth tyres over them.

[IES 2010]

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true but R is not a correct explanation of A.
  3. A is true but R is false. 
  4. A is false but R is true.
Answer: Option B
Explanation:
  • Smooth tyres offer high friction on dry pavements due to larger contact area.
  • New tyres offer high friction on wet pavements due to the presence of treads.

5.  Match List-I with List-II and select the correct answer using the code given below the lists:

List-I

  1. Lateral friction
  2. Cut-off lagoons
  3. Skid
  4. Sight distance

List-II

  1. Disparity between relevant travel distances
  2. Vehicle movement on a curve
  3. Summit curves
  4. Prevention of flooding

[IES 2011]

Codes: A B C D

  1.  2  1  4  3
  2.  3  1  4  2
  3.  2  4  1  3
  4.  3  4  1  2
Answer: Option C
Explanation:
  • The lateral friction between the tyres and road surface helps to maintain the stability of a vehicle while moving on a horizontal curve.
  • Cut-off lagoons are the structures created to prevent the flooding of roads.
  • Skidding occurs when the linear distance travelled by the wheels is more than the circumferential distance travelled by them.
  • The design of summit curve depends on the sight distance to be provided.

6.  Assertion (A): IRC has recommended a minimum coefficient of friction in the longitudinal direction on wet pavements after allowing a suitable factor of safety in the range 0.15 – 0.30.

Reason (R): When the longitudinal coefficient of friction of 0.40 is allowed for stopping the vehicle, the resultant retardation is 3.93 m/sec2.

[IES 2011]

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true and R is not a correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Answer: Option D
Explanation:
The IRC recommended value of coefficient of longitudinal friction is 0.35 - 0.40.

7.  What is the value of the resultant retardation in m/s2 when a longitudinal friction coefficient of 0.4 is allowed for stopping the vehicle on road ?

[IES 2013]

  1. 0.98
  2. 1.95
  3. 2.93
  4. 3.93
Answer: Option D
Explanation:
The average skid resistance coefficient (f) can be calculated using the below formulae:
f =
a / g
, 
2L / gt2
, 
v2 / 2gL

Using the expression, f =
a / g

⇒ 0.4 =
a / 9.81

⇒ a = 3.93 m/s2

8.  If a vehicle travelling at 40 kmph was stopped within 1.8 sec after the application of the brakes, then the average skid resistance coefficient is _____ .

[IES 2014]

  1. 0.63
  2. 0.73
  3. 0.83
  4. 0.93
Answer: Option A
Explanation:
Initial speed = 40 kmph = 11.11 m/s
Final speed = 0
Time taken, t = 1.8 s
Now, Retardation, a =  
Change in speed / t

⇒ a =  
11.11 - 0 / 1.8
= 6.173 m/s2
Thus, average skid resistance coefficient, f =  
a / g
=  
6.173 / 9.81

⇒ f = 0.629 ≈ 0.63

9.  A four-lane divided highway, with each carriageway being 7.0 m wide, is to be constructed in a zone of high rainfall. In this stretch, the highway has a longitudinal slope of 3% and is provided with a camber of 2%. What is the hydraulic gradient on this highway in this stretch ?

[IES 2015]

  1. 4.0%
  2. 3.6%
  3. 4.5%
  4. 3.0%
Answer: Option D
Explanation:
The longitudinal slope provided on the highway is the hydraulic gradient of the highway. Hence, hydraulic gradient is 3.0%

10.  In an area of heavy rainfall, a State Highway of high-type bituminous surface with four lanes (14.0 m wide) is to be constructed. What will be the height of the crown of the road relative to the edges for a composite camber (i.e. middle half as parabola and the rest as straight lines) ?

[IES 2015]

  1. 14 cm
  2. 21 cm
  3. 28 cm
  4. 7 cm
Answer: Option A
Explanation:
  • The height of the crown will remain the same irrespective of the camber profile.
  • In case of areas of heavy rainfall and for a State Highway, camber to be provided = 2%
  • The height of crown =
    W / 2
    x camber =
    14 / 2
    x
    2 / 100

    Thus, the height of crown = 0.14 m (or 14 cm)
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