Railway Engineering MCQ Quiz in मल्याळम - Objective Question with Answer for Railway Engineering - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Apr 14, 2025
Latest Railway Engineering MCQ Objective Questions
Top Railway Engineering MCQ Objective Questions
Railway Engineering Question 1:
If the track is laid on the place in a curve of 5 degree, the allowable ruling gradient on the curve is(Ruling Gradient = 1 in 200 )
Answer (Detailed Solution Below)
Railway Engineering Question 1 Detailed Solution
Concept:
Grade compensation:
The ruling gradient is the maximum gradient on a particular section, but if a curve lies on a ruling gradient, the resistance due to gradient is increased by that due to curvature, and this further increases the resistance beyond the ruling gradient. In order to avoid resistances beyond the allowable limits, the gradients are reduced on curves, and this reduction is known as the grade compensation for the curve.
Allowable ruling gradient = Ruling gradient - Grade compensation
Grade compensation (GC) for BG = 0.04% per degree of curve
Grade compensation (GC) for MG = 0.03% per degree of curve
Grade compensation (GC) for NG = 0.02% per degree of curve
Calculation
Given,
Degree of curve D = 5°
Ruling Gradient = 1 in 200
So for 5° curve, compensation,
= 0.04 × 5 = 0.2%
∴ Ruling gradient = (1/200) × 100 = 0.5%
∴ Allowable gradient = (0.5 - 0.2)% = 0.3/100 = 1/333.33
∴ Allowable Gradient = 1 in 334
Railway Engineering Question 2:
An electric locomotive running at 60 kmph on a curved track of 1.68 m gauge laid at 800 m radius should be provided with superelevation of the rail by an amount of
Answer (Detailed Solution Below)
Railway Engineering Question 2 Detailed Solution
Concept:
Superelevation:
To counteract the effect of centrifugal force, the level of the outer rail is raised above the inner rail by a certain amount to introduce the centripetal force.
This raised elevation of the outer rail above the inner rail at a horizontal curve is called superelevation.
For any gauge super elevation ‘e’ (in m) is given as
\(e = \frac{{{\rm{GV}}^2}}{{127{\rm{R}}}}\)
Where, G = Gauge distance
V = velocity in Kmph
R = Radius of curve
Calculation:
Given,
G = Gauge distance = 1.68 m
V = velocity in Kmph = 60
R = Radius of curve = 800 m
\(e = \frac{{{\rm{GV}}^2}}{{127{\rm{R}}}}\)
\(e = \frac{{1.68 \times {{60}^2}}}{{127 \times 800}}\) = 59.52 mmRailway Engineering Question 3:
The distance between the gauge faces of the stock rail the tongue rail at the heel of the switch is called
Answer (Detailed Solution Below)
Railway Engineering Question 3 Detailed Solution
Concept
Heel Clearance or Heel Divergence
- It is distance between running face of stock rail and tongue rail at heel of switch.
- Indian standard specifies heel clearance for BG track as 13.7 cm – 13.9 cm.
Flange way Clearance
- The distance between the adjacent faces of the running rail and the guard or the check rails provides clearance for passage of wheel flanges.
Throw of Switch
- The gap through which the toe of the gauge rail moves sideways to provide a path for the required direction over the turnout.
- Throw of switch according to Indian standard: BG = 9.5 cm, MG/NG = 8.9 cm
Switch angle
- The angle between the gauge face of stock rail and tongue rail.
Railway Engineering Question 4:
At points and crossings, the total number of sleepers for 1 in 12 turnouts in broad gauge is
Answer (Detailed Solution Below)
Railway Engineering Question 4 Detailed Solution
Explanation:
(i) Total number of the sleeper at point and crossing depend upon the turnout and in indian railway generally, two types of turnout are provided i.e. 1 in 12 and 1 in 8.5
(ii) For 1 in 12 turnouts, the total number of sleeper = 70
(iii) For 1 in 8.5 turnouts, the total number of sleeper = 62
Railway Engineering Question 5:
Composite Sleeper Index’ is employed to determine
Answer (Detailed Solution Below)
Railway Engineering Question 5 Detailed Solution
Concept:
This is an index to determine the suitability of a particular timber for use as a sleeper.
Note:
This index measures the mechanical strength of timber, derived from its composite properties of
strength and hardness.
The C.S.I. is calculated from the relation.
C.S.I = \(\frac{{{\rm{S}} + 10{\rm{H}}}}{{20}}\)
Where,
S = Strength Index, both for green and dry timber at 12% moisture content.
H = Hardness Index, both green and dry timber at 12% moisture content.
Minimum value of C.S.I. prescribed on Indian Railways is as follows:
Type of sleeper | Minimum C.S.I |
Track Sleeper | 783 |
Crossing Sleeper | 1352 |
Bridge Sleeper | 1455 |
Railway Engineering Question 6:
The ballast material generally used in Indian railways consists of
Answer (Detailed Solution Below)
Railway Engineering Question 6 Detailed Solution
Explanation:
The following materials for Ballast can be used on the railway track.
1. Broken Stone
2. Gravel
3. Cinders / Ashes
4. Sand
5. Kankars
6. Moorum
7. Brick Ballast
Among above materials, broken stone from Igneous rocks like quartzite and granite forms the excellent ballast materials. When these are not available then lime stone and sand stone can also be used as good ballast material.
Some of the main functions of ballast are followings:
a) To provide firm and level bedded foundation for the sleepers and rails to rest on
b) To protect the surface of subgrade and to form an elastic bed
c) To transmit and distribute the loads from the sleepers to the subgrade
d) To allow for maintaining correct track level without disturbing the rail road bed
e) To hold the sleepers in position during the passage of trains
f) To provide lateral stability to the track as a whole.
Railway Engineering Question 7:
Calculate the number of sleepers required for 1 km railway track, if sleeper density is (n + 4) for broad gauge and the length of one rail for a broad gauge is 13 m.
Answer (Detailed Solution Below)
Railway Engineering Question 7 Detailed Solution
Concept:
Sleeper density is denoted as n + x.
Here n + x denotes the number of sleepers to be used for one rail length.
n = length of one rail.
Calculation:
Given Data:
Sleeper density = n + 4
Length of one rail = 13 m ⇒ n + 4 = 17
⇒ 17 sleepers are required for 13 m
For 1 m, number of sleepers required \(= \frac{{17}}{{13}}\) = 1.308
For 1 km, the number of sleepers required \(= \frac{{17}}{{13}} \times 1000 = 1307.69\) ≈ 1308 Numbers.Railway Engineering Question 8:
For prevention of creep in a railway track, which of the following methods can be adopted?
i. Using creep indicator
ii. Using anchors below the rails
iii. Use of steel sleepers
iv. Avoid fast movement of trains
Answer (Detailed Solution Below)
Railway Engineering Question 8 Detailed Solution
Explanation:
Creep:
- Creep is defined as the longitudinal movement of rail with respect to sleepers.
- Rails have a tendency to gradually move in the direction of dominant traffic.
Remedial measures to prevent creep of rails
- Pulling of anchors and anti-creepers should be done.
- The number of sleepers per rail length should be increased.
- Pulling back the rail to its original position.
- One should provide adequate ballast and packing carefully.
- For good grip, steel sleepers are used.
Railway Engineering Question 9:
The distance between two rails in a narrow railway gauge is ____.
Answer (Detailed Solution Below)
Railway Engineering Question 9 Detailed Solution
The correct answer is 762 mm.
- In a narrow-gauge system, the distance between two tracks is 762 mm and 2 ft 6 inches.
Key Points
- The gauge of a railway track is defined as the clear minimum perpendicular distance between the inner faces of the two rails.
- The narrow gauge is also called a small gauge or small line.
- In 2015 there was a 1500 km narrow-gauge route which is considered to be about 2% of the total Indian Rail network.
- The Darjeeling Mountain Railway was declared UNESCO world heritage in 1999.
- Kalka Shimla Railway is a narrow gauge.
Additional Information
- a railway track with a smaller distance between the lines than the standard gauge of 56 1/2 in.
- The different gauges can broadly be divided into the following four categories:
- Broad Gauge width 1676 mm to 1524 mm or 5'6” to 5'0”
- Standard Gauge width 1435 mm and 1451 mm or 4'-8⅟2”
- Metre Gauge: width 1067 mm, 1000 mm, and 915 mm or 3'-6”, 3'-33/8” and 3'-0”
- Narrow Gauge: width 762 mm and 610 mm or 2'-6” and 2'-0”.
Railway Engineering Question 10:
Flat, gravitational and hump yard are the types of:
Answer (Detailed Solution Below)
Railway Engineering Question 10 Detailed Solution
Explanation:
A Yard can be defined by the functions it performs. It is the place or the activity center on a Railway system, where the trains or rolling stock or group of rolling stocks are received, reformed into trains or loads after marshalling and are dispatched to their destinations.
Classification of yards: 1. Marshalling Yard 2. Passenger yard 3. Goods yard 4. Coaching yards
Marshalling yard:
- The arrangement made with an elaborate set of sidings for sorting of wagons at important stations according to traffic requirements, is called marshalling yard.
- Goods tarins are marshaled so that wagons are placed in the order of the stations at which these are to be detached.
- Marshalling Yard can be classified under three groups on the basis of the method of sorting out trains:
- Flat Yard
- Hump Yard
-
Gravity Yard
Locomotives yard:
- This is the yard which houses the locomotive. Facilities for watering, fueling, examining locomotives, repairing, etc., are provided in this yard.
-
The yard layout is designed depending upon the number of locomotives required to be housed in the locomotive shed.
Coaching yard:
- The main function of a coaching yard is to deal with the reception and dispatch of passenger trains.
-
Depending upon the volume of traffic, this yard provides facilities such as watering and fueling of engines, washing of rakes, examination of coaches, charging of batteries, and trans-shipment of passengers.
Passenger yard:
- The main function of passenger yards is to provide facilities for the safe movement of passengers and passenger bogies.
- Passenger platforms are used as passenger yards.