Module-94C: Importance of Surveying, Principles and Classification, Mapping Concepts, Coordinate System, Map Projections
Module-94C: Importance of Surveying, Principles and Classification, Mapping Concepts, Coordinate System, Map Projections
Module-94C: Importance of Surveying, Principles and Classification, Mapping Concepts, Coordinate System, Map Projections
MODULE- 94C
Importance of surveying, principles and
classification, mapping concepts, coordinate
system, map projections
______________________________________
1. The local mean lime at a place located in longitude 90 40' E when
the standard time is 6 hours and 30 minutes and the standard
meridian is 82° 30' E is
Ans:
a) Geodetic surveying
b) Plane surveying
c) Preliminary surveying
d) Topographical surveying
Ans: Geodetic surveying
a) 7 hr 00 m
b) 8 hr 00 m
c) 8 hr 30 m
d) 9 hr 00 m
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SUB: SURVEYING MODULE-94C
Ans:
a) 1: 24000
b) 1: 62500
c) 1: 100000
d) 1: 500000
Ans:
Large scale maps are better for showing individual buildings in
detail because they only cover a small area of land. So 1:
24000 is largest scale.
a) 1/3 hr
b) 1/3
c) 1/20
d) 200 sec
Ans:
1 degree = 60 minutes
1 minute = 60 sec.
So 20 minutes =
6. You need to find the scale of a map have been given for a class
project. In order to do so, you decide to measure a distance
between two real world features and relate that to the distance
shown on the map. You find that the distance between two
buildings is 400 feet. On the map that same distance is 3 inches.
What is the scale of the map?
a) 1: 1200
b) 1: 16000
c) 1: 1600
d) 1: 100
Ans:
3 inches : 400 feet
3 inches : inches
1: 1600
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SUB: SURVEYING MODULE-94C
a) A cartogram
b) Equal area
c) Proportional symbol
d) Topographic
Ans:
A cartogram is a map in which some thematic mapping
variable – such as travel time, population is substituted for
land area or distance. The geometry or space of the map is
distorted in order to convey the information of this alternate
variable. There are two main types of cartograms: area and
distance cartograms
a) 33 miles
b) 50 miles
c) 69 miles
d) 111 miles
Ans:
Radius of earth= 6370 km = 6370 / 1.609 = 3960 miles
Perimeter = 3960 = 24875 miles
Distance for 360 degrees = 24875 miles
Distance for 1 degree = 24875/360 = 69 miles.
10. A half circle of 180 degrees of arc which connects the earth's poles
is known as a:
a) Latitude
b) Meridian
c) Longitude
d) Parallel
Ans: Meridian
a) Latitude
b) Longitude
c) Parallel
d) Meridian
Ans: Latitude
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SUB: SURVEYING MODULE-94C
13. General purpose maps that depict the shape and elevation of
terrain, and usually portray the surface features of relatively small
areas, are known as:
a) Topographic maps
b) Thematic maps
c) Mercator maps
d) Value-by-area maps
14. Which type of map projection shows true directions from one
central point to all other points?
a) Equivalent
b) Conformal
c) Equidistant
d) Azimuthal
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SUB: SURVEYING MODULE-94C
Ans. Azimuthal
16. The Prime Meridian runs through the Royal Observatory at:
17. If a map is called conformal, then it is correct for areas in terms of:
a) Area
b) Shape
c) Distance
d) Direction
Ans: Shape.
Conformal maps preserve both angles and the shapes of
infinitesimally small figures, but not necessarily their size or
curvature.
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SUB: SURVEYING MODULE-94C
A) Equidistant
B) Geoidal
C) Conformal
D) Azimuthal
Ans: Geoidal
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SUB: SURVEYING MODULE-95C
MODULE- 95C
a) b)
c) d)
Ans:
=330
a) b)
c) d)
Ans:
By observation FB and BB of DE are differ by
∴The points D and E are free from local attraction
∴The FB of EA is correct and is
The BB of EA =
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SUB: SURVEYING MODULE-95C
Correction=
The FB of AB =
The BB of AB =
Correction =
The FB of BC=
a) to b) to
c) to d) to
a) b)
c) d)
Ans:
True bearing =
=
Magnetic bearing =
=
W.C.B=360-
=
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SUB: SURVEYING MODULE-95C
a) 10000 b) 6561
c) 1000 d) 1656
Ans:
( )
6. The plan of a map was photo copied to a reduced size such that a
line originally 100mm, measures 90mm. the original scale of the
plan was 1:1000. The revised scale is
a) 1:900 b) 1:1111
c) 1:1121 d) 1:1221
Ans:
Originally 1 mm in plan = 1000 mm actual
100mm in plan=1000×100mm
Now 90mm in plane =1000×100mm
1mm=1000×
a) b)
c) d)
Ans:
Magnetic bearing
True bearing
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SUB: SURVEYING MODULE-95C
a) 1 b)
c) d)
Ans:
P(100 N, 200 E) E
a) 273.205,938.186
b) 273.205, 551.815
c) 551.815, 551.875
d) 551.815,938.186
Ans:
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SUB: SURVEYING MODULE-95C
a) b)
c) d)
Ans:
Surveyor used the scale 100m to 1cm
Distance between station on map =3500/100=35cm
As the actual scale of map is 50m to 1cm
True distance on the ground =35×50=1750m
11. An old map was plotted to a scale of 40m to 1cm. over the years,
this map has been shrinking, and a line originally 20cm long is only
19.5cm long at present. Again 20m chain was 5cm too long. If the
present area of the map measured by plain meter is 125.50 find
the true area of the land.
a) b)
c) d)
Ans:
Scale: 1 cm =40 m
Area on ground =132.0184
( )
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Sub: SurveyingTopic :Levelling
MODULE- 96C
Levelling
________________________________________________________
R.L of B =100-2.105-1.105
=96.79m
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Sub: SurveyingTopic :Levelling
P 2.800m 1.700m
Q 2.700m 1.800m
is close to P and is close to Q. If the reduced level of station P
is 100.000m, the reduced level of station Q is
a) 99.00m b) 100.00m
c) 101.00m d) 102.00m
Ans:
5. A bench mark (BM) with reduced level (RL) =155.305m has been
established at the floor of a room. It is required to find out the RL
of the underside of the roof (R) of the room using spirit leveling.
The back sight (BS) to the BM has been observed as 1.500m where
as fore sight (FS) to R has been observed as 0.575(staff held
inverted) the RL(m) of R will be
a) 155.88 b) 156.23
c) 157.38 d) 157.86
Ans:
=157.380
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Sub: SurveyingTopic :Levelling
a) 0.025cm b) 0.25cm
c) 2.5cm d) 5.0cm
Ans:
Ans:
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Sub: SurveyingTopic :Levelling
0.067269
a) 51.197, 3.103
b) 50.197,3.103
c) 51.197,2.103
d) 52, 3
Ans:
51.197
54.3-51.197=3.103m
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Sub: SurveyingTopic :Levelling
11. Reciprocal levelling was done between two points A and B situated
on the opposite sides of a valley 730m wide.
Instrument height of staff at staff
Instrument Height of Staff Staff
At Instrument at reading
A 1.463 B 1.688
B 1.463 A 0.991
Determine the difference in level between A and B and the amount
of collimation error.
a) 0.3485, 0.1235 b) 0.2485, 0.1235
c) 0.2485, 0.2235 d) 0.1, 0.2
Ans:
In correct level difference between A&B in 1st case=1.463-
1.688=-0.225
Incorrect level difference between A&B in 2nd case=0.991-
1.463=0.472
True level difference
13. Calculate the combined correction for curvature and refraction for
distance of 5Km and 500m
a) 1.682,0.168
b) 1.682,0.0168
c) 0.168,1.682
d) 0.0168,1.682
Ans:
( ) ( ) is for refraction
0.0673
∴0.0673
0.0673
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Sub: SurveyingTopic :Levelling
15. A back sight reading on B.M= 100m was 3.25m. The inverted staff
reading to the bottom of girder was 1.25m. the RL of the bottom of
girder is
a) 101.25 b) 102.0
c) 104.50 d) 103.25
Ans:
100+3.25+1.25=104.50
16. A light house of 120m height is just visible above the horizon from
a ship. The correct distance(m) between the ship and the light
house considering combined correction for curvature and refraction
is
a) 39.098 b) 42.226
c) 39098 d) 42226
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Sub: SurveyingTopic :Levelling
Ans:
0.0673
0.0673
D=42.226Km or 42226 m
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Sub: SurveyingTopic : Theodolite
MODULE- 97C
Theodolite traversing
________________________________________________________
( )( ) ( )
d=128.746 m
∴h=128.746(tan5+tan0.5)
=12.387 m
Ans:
R.L of B= 880.88+1.1-237.7-2.5=641.78
Correction for curvature =
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Sub: SurveyingTopic : Theodolite
Refraction= (-) ( )
0.0673
a) 396.86m b) 396.79m
c) 396.05m d) 396.94m
Ans:
18.917 299.403
-348.45 65.325
-197.97 -404.113
197.59 -181.058
0.8131x - 329.917 0.556x – 220.443
x = 396.79 m
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Sub: SurveyingTopic : Theodolite
Ans:
Latitude Departure
200 0
707.1 707.1
- 907 0
0 707.1
∴ =0 = -707.1
= 707.1 ∴ θ = 270
= +
= –1 or =
=
∴ = 200(1 + 500)
= 501 × = 100.2 m
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Sub: SurveyingTopic : Theodolite
Ans:
∑ = 501, ∑ = 499, ∑ =499, ∑ =501
p +q = 100
p -q = -50
0.9396 p + 0.4628 q = 100
0.342 p – 0.886 q = -50
P = 66.07
∴ x = 300 + 66.07 = 362
∴ y = 200 + 66.07 = 222.59
8. A tachometer fixed with stadia wires 4mm apart has its object glass
(f=200mm) fixed at a distance of 250mm from the trunnion axis.
The tachometric distance equation is
a) D=100S+0.45
b) D=50S+0.45
c) 100S+0.25
d) 100S+0.35
Ans:
= + or = 1+ = -1
= = –1
( )
+ d = D = . S +(f +d)
9. Two tangents spaced 6.0m apart were fixed on a sub tense bar and
the vertical angles measured on the two upper and lower targets
were and respectively. If the lower target was
at an elevation of 249.2m what was the height of instrument
a) 240.45 b) 239.45
c) 241.45 d) 242.45
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Sub: SurveyingTopic : Theodolite
Ans:
=
=
∴ 0.0435 =
0.0258 =
∴
∴6 0.0258 + 0.0258V = 0.0435V
V = 8.746m
d = 339 m
Level of collimation = 249.2 – 8.746
= 240.454
10. The constants for an instrument are 1000 and 0.5.calculate the
distance from the instrument to the staff when the micrometer
readings are 3.246 and 5.246. the staff intercept is 2.0m and
vertical angle measured is , the staff being vertical
a) 189.95 b) 188.95
c) 190 d) 200
Ans:
D= .S +C.
= .2× + 0.5 ×
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SUB: SURVEYING MODULE-98C
MODULE- 98C
Plane Table surveying
______________________________________
1. During orientation of a plane table
a) The farthest point is sighted
b) The nearest point is sighted
c) Either (a) or (b)
d) The previous station is sighted
a) Radiation
b) Traversing
c) Intersection
d) Resection
Ans: Intersection
a) Inaccurate centering
b) Long sight
c) Shrinkage of drawing sheet
d) Few observations
5. Lehmann’s rule states that while looking towards the station the
plane table location to be fixed is always to
a) Left of each of the rays
b) Right of each of the rays
c) Left or right of the ray to the most distant station
d) Left or right of each of the rays
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SUB: SURVEYING MODULE-98C
a) A trough compass
b) Back sighting
c) Observations of two well-defined points
d) Observation of three well-defined points
a) Great triangle
b) Great circle
c) Centre of great circle
d) Orthocenter of great triangle
a) a back ray
b) trough compass
c) three point problem
d) two point problem
10. The instrument used for accurate centering in plane table survey is
a) Spirit level
b) Alidade
c) Plumbing fork
d) Trough compass
a) Radiation
b) Intersection
c) Resection
d) Traversing
Ans: Resection
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SUB: SURVEYING MODULE-98C
12. The resection by two point problem as compared to three point problem
13. The two point problem and three point problem are methods of
a) Resection
b) Orientation
c) Traversing
d) Resection and orientation
Ans: Resection
a) Tacheometry
b) Trigonometrical leveling
c) Plane table surveying
d) Theodolite surveying
a) Radiation
b) Intersection
c) Traversing
d) Resection
Ans: Traversing
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SUB: SURVEYING MODULE-99B
MODULE- 99C
Errors and Adjustments
______________________________________
1. One of the characteristics of an random errors is that
Ans: Small errors occur more frequently than the large errors
a) 0.01 m
b) 0.02 m
c) 0.04 m
d) 0.16 m
Ans:
Ans: Precision
a)
b)
c)
d)
Ans:
∫
√
=
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SUB: SURVEYING MODULE-99B
a) Gross errors
b) Systematic errors
c) Random errors
d) All the above
a) 6.745m
b) 20m
c) 10m
d) 0.6745m
Ans:
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SUB: SURVEYING MODULE-99B
a)
b)
c)
d)
Ans:
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SUB: SURVEYING MODULE-100C
MODULE- 100C
Curves
______________________________________
= 80.4m
a) 310.4 b) 320.4
c) 330.4 d) 190.4
Ans:
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SUB: SURVEYING MODULE-100C
= ∴β=32.146
or 180 – 32.1467
=147.853
∴γ =180-30-147.8533
= =2.146
∴ R = 320.4m
a) 490,2997,2758
b) 490, 2758, 2997
c) 290, 2997, 2758
d) 290, 2758, 2997
Ans:
R sec 14 – R =15
∴ R = 490 m
PQ = 490 × = 122.17
Length of curve = 490 × ×
=239.46m
Chainage at P= 2880 - 122.17
= 2757.83
Chain age at R = 275783 + 239.46
=2997.29 m
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SUB: SURVEYING MODULE-100C
Ans:
PQ = 300 = 97.476m
PQ = 300× × = 188.496 m
Chain age at P = 1022 – 97.476
=924.524
Chain age at R = 924.524 +188.496
= 1113.02
a) 3800 m b) 3803 m
c) 3806 m d) 3809 m
Ans:
= ∴ = 14.4477
Β=78–14.477=63.522
γ = 66. or 113.78
∴ =
∴ 3803 m
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SUB: SURVEYING MODULE-100C
a) 640m b) 320m
c) 160m d) 80m
Ans:
FG = 20 m
CD = 180.2 m
R +R
= 180.2
∴ R = 640 m
Ans:
= 720 m
720 = R +R + 2R
= 0.79 = 0.613
θ=
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SUB: SURVEYING MODULE-100C
( ( ))
( )
8. If the degree of curve is and if the chain length is 30m, then the
radius of curve is equal to
a) 5400m b) 1720m
c) d) m
Ans:
R × 10 × = 30
R= = 1718.8m so 1720 m
a) b)
c) d)
Ans:
I=
R = 2. R
∴ =2
∴ =
= ∴ ∆ = 12
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SUB: SURVEYING MODULE-100C
10. The ratio of the apex distance of a curve of and radius R deflecting
through is
a) sec b) 1-sec
c) Cot d)
Ans:
[ ]
= sec 1
11. The total deflection angle for a simple circular curve at a point
is . A tangent is drawn which as a length of 150m between the
two tangents and this makes an angle of with the back tangent.
The radius of curve is
a) 181m b) 451m
c) 305m d) 498m
Ans:
150 = R ×2
∴ R = 181 m
12. If the radius of simple circular curve is 400m and deflection angle
is , the mid ordinate is
a) 100m b) 200m
c) 400m d) 800m
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SUB: SURVEYING MODULE-100C
Ans:
Mid ordinate = R - R
∴ 400 - 400 = 200m
13. The radius of a simple circular curve is 30m and the length of the
specified chord is 30m. the degree of the curve is
a) 57.29 b) 3.7
c) 55.6 d) 37.03
Ans:
30 θ = 30
∴θ = = =
14. For a curve of radius 100m and normal chord 10m. deflection
angle=
a) b)
c) d)
Ans:
∴∆=
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SUB: SURVEYING MODULE-100C
a) ⁄ b) 180-
c) 180+ d) 90+
Ans:
Intersection angle
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