Highway Engineering PYQ – TN TRB Lecturer
31. PSI stands for _____ .
[TN TRB 2017: 1 Mark]
- Pavement Serviceability Index
- Permanent Serviceability Index
- Pavement Service Index
- Present Serviceability Index
Explanation:
PSI stands for Present Serviceability Index. Present Serviceability Index (PSI) value is used for the functional evaluation of the pavements.
32. The allowable centrifugal ratio for railways is _____ .
[TN TRB 2021: 1 Mark]
- 1/6
- 1/4
- 1/8
- 1/10
Explanation:
- Centrifugal ratio for railways is 1/8.
- Centrifugal ratio for highways is 1/4.
33. Find the radius of curvature of the bubble tube if the length of 1 division is 2 mm and if the angular value of 1 division is 1 minute.
[TN TRB 2021: 1 Mark]
- 6.87 m
- 0.033 m
- 687 m
- 68.7 m
Explanation:
Sensitivity of bubble tube (in seconds),
α =
Radius of curvature of bubble tube
x 206265
⇒ 1’ x 60” =
Radius of curvature of bubble tube
x 206265
⇒ Radius of curvature of bubble tube = 6875.5 mm (or) 6.876 m
34. The maximum length of offset is limited to _____ .
[TN TRB 2021: 1 Mark]
- 3 m
- 30 m
- 15 m
- 20 m
Explanation:
35. Which of the following Indian Standard is used to find the softening point of bitumen ?
[TN TRB 2021: 1 Mark]
- IS 1206 – Part 2
- IS 1206 – Part 3
- IS 1203
- IS 1205
Explanation:
- IS 1206 (Part II): 1978 – Methods For Testing Tar and Bituminous Materials: Determination of Viscosity Part II – Absolute Viscosity
- IS 1206 (Part III): 1978 – Methods For Testing Tar and Bituminous Materials: Determination of Viscosity Part III – Kinematic Viscosity
- IS 1203:1978 – Methods For Testing Tar and Bituminous Materials: Determination of Penetration
- IS 1205:1978 – Methods For Testing Tar and Bituminous Materials: Determination of Softening Point
36. A vehicle moving at 60 kmph on a bituminous dry surface is suddenly brought to rest by braking. The coefficient can be assumed to be 0.5. Calculate the distance over which the vehicles comes to a stop.
[TN TRB 2021: 1 Mark]
- 30.12 m
- 28.3 m
- 2.83 m
- 0.3 km
Explanation:
Speed of vehicle , v = 60 Kmph (or) 16.67 m/s
Braking distance =
2gf
=
2 x 9.81 x 0.5
⇒Braking distance = 28.3 m
37. If q be the angle of slope and L be the sloping distance the correction for slope is given by _____ .
[TN TRB 2021: 1 Mark]
- L (1 – cos ϴ) (negative)
- L (cos ϴ – 1) (negative)
- L (1 – cos ϴ) (positive)
- L (cos ϴ -1) (positive)
Explanation:
The correction for slope is given by,
38. The application of a bituminous binder to an existing surface to ensure a bond is called as _____ .
[TN TRB 2021: 1 Mark]
- Prime coat
- Seat coat
- Surface Pressing
- Tack coat
Explanation:
39. Determine the capacity of a single lane (unidirectional) on a rural highway in India. For a design speed of 50 kmph the average length of car can be taken as 5 m. The perception-brake reaction time can be taken to be 2.5 sec. The coefficient of friction can be assumed to be 0.5.
[TN TRB 2021: 2 Marks]
- 845 vehicles per hour per lane
- 847 vehicles per hour per lane
- 833 vehicles per hour per lane
- 840 vehicles per hour per lane
Explanation:
Since, reaction time is greater than 2 secs,
Space Headway,
S = 0.278 Vt +
2gf
+ L
⇒ S = (0.278 x 50 x 2.5) +
2 x 9.81 x 0.5
+ 5
⇒ S = 59.45 m
Theoretical capacity, C =
S
x V =
11.95
x 50
⇒ C = 841.11 veh per hour
The closest option is 840 veh per hour.
40. A level is set up at a point 150 m from A and 100 m from B, the observed staff readings at A and B are 2.525 and 1.755 respectively. Find the true differences of level between A and B.
[TN TRB 2021: 2 Marks]
- 0.7762
- 0.7792
- 0.7692
- 0.7612
Explanation: