Surveying Multiple Choice Questions - Group 2
Q.1. The minimum number of satellites needed for a GPS to determine its position precisely is ____
GATE-2016
2
3
4
24
GPS receivers require timing codes from at least four GPS satellites to fix a position.
Three satellites are used to trilaterate, and a fourth satellite's timing code is used as a timing reference; the position can be fixed with three timing codes, but the clock precision and synchronization requirements make using a fourth satellite's timing code much more practical.
There fore correct answer is Option - 3.
Three satellites are used to trilaterate, and a fourth satellite's timing code is used as a timing reference; the position can be fixed with three timing codes, but the clock precision and synchronization requirements make using a fourth satellite's timing code much more practical.
There fore correct answer is Option - 3.
Q.2. The system that uses the Sun as a source of electromagnetic energy and records the naturally
radiated and reflected energy from the object is called______
GATE-2016
Geographical Information System
Global Positioning System
Passive Remote Sensing
Active Remote Sensing
There are two types of remote sensing instruments—passive and active. Passive instruments detect natural energy that is reflected or emitted from the observed scene. Passive instruments sense only radiation emitted by the object being viewed or reflected by the object from a source other than the instrument. Reflected sunlight is the most common external source of radiation sensed by passive instruments.
There fore correct ans. is Option - 3.
There fore correct ans. is Option - 3.
Q.3. The staff reading taken on a workshop floor using a level is 0.645 m. The inverted staff reading
taken to the bottom of a beam is 2.960 m. The reduced level of the floor is 40.500 m. The reduced
level (expressed in m) of the bottom of the beam is______
GATE-2016
44.105.
43.460.
42.815.
41.145.
R. L. of workshop floor = 40.500 m.
Staff reading on workshop floor = 0.645 m.
H. I. = 40.500 + 0.645 = 41.145 m
The inverted staff reading taken to the bottom of a beam is 2.960 m. There fore this reading is added in H. I.
R. L. of bottom of the beam = 41.145 + 2.960 = 44.105 m
There fore correct answer is Option - 1.
Staff reading on workshop floor = 0.645 m.
H. I. = 40.500 + 0.645 = 41.145 m
The inverted staff reading taken to the bottom of a beam is 2.960 m. There fore this reading is added in H. I.
R. L. of bottom of the beam = 41.145 + 2.960 = 44.105 m
There fore correct answer is Option - 1.
Q.4. In a region with magnetic declination of 20E, the magnetic Fore bearing (FB) of a line AB was
measured as N79050'E. There was local attraction at A. To determine the correct magnetic bearing
of the line, a point O was selected at which there was no local attraction. The magnetic FB of line
AO and OA were observed to be S52040'E and N50020'W, respectively. What is the true FB of line
AB?
GATE-2015
N81050'E.
N82010'E.
N84010'E.
N77050'E.
First convert all readings in to WCB
Station O is free form local attraction hence start form this station.
Correct FB of OA = 309040'
Correct BB of AO = 309040' + 1800 - 3600 = 129040'
Observed BB of AO = 127020'
error = Observed reading - Correct reading = 127020' - 129040'
error = - 2020'
correction = + 2020'
Correct FB of AB = 79050' + 2020' = 82010'
Declination of 20E there for Correct FB of AB = 82010' + 20 = 84010'
in RB = N84010'E
There fore correct answer is Option - 3.
| Line | RB | WCB |
| AB | N79050'E | 79050' |
| AO | S52040'E | 127020' |
| OA | N50020'W | 309040' |
Correct FB of OA = 309040'
Correct BB of AO = 309040' + 1800 - 3600 = 129040'
Observed BB of AO = 127020'
error = Observed reading - Correct reading = 127020' - 129040'
error = - 2020'
correction = + 2020'
Correct FB of AB = 79050' + 2020' = 82010'
Declination of 20E there for Correct FB of AB = 82010' + 20 = 84010'
in RB = N84010'E
There fore correct answer is Option - 3.
Q.5. A rod 5 km length is shown by a straight line of 10 cm on a Index plan. What is R. F. of Index plan?
1:5000
1:500
1:50
none of the above
10 cm on paper = 5 km on ground
R. F. = Distance on paper/Distance on ground
R. F. = 10 cm/5*1000*100 cm
R. F. = 1/50000 cm
R. F. = 1:50000
There fore correct answer is Option - 4.
R. F. = Distance on paper/Distance on ground
R. F. = 10 cm/5*1000*100 cm
R. F. = 1/50000 cm
R. F. = 1:50000
There fore correct answer is Option - 4.
Q.6. There are ______ links are in one meter chain.
4
3
5
1
Length of one link = 20 cm.
∴ No.s of links in 1 m = 1 m /0.2 m = 5 No.s
There fore correct answer is Option - 3.
∴ No.s of links in 1 m = 1 m /0.2 m = 5 No.s
There fore correct answer is Option - 3.
Q.7. Match the following related to surveying.
| Group - I | Group - II |
| P) Alidade | 1) Chain Survey |
| Q) Arrow | 2) Levelling |
| R) Bubble tube | 3) Plant table surveying |
| S) Stedia hair | 4) Theodolite |
GATE-2014
P – 3, Q – 2, R – 1, S – 4
P – 2, Q – 4, R – 3, S – 1
P – 1, Q – 2, R – 4, S – 3
P – 3, Q – 1, R – 2, S – 4
In above table Group - I consist of list of instruments used in survey work. And Group - II consist of there uses for particular survey purpose.
i.e. Alidade > used in Plant table surveying
Arrow > used in Chain Survey
Bubble tube > used in Levelling
Stedia hair > used in Theodolite
There fore correct answer is Option - 4.
i.e. Alidade > used in Plant table surveying
Arrow > used in Chain Survey
Bubble tube > used in Levelling
Stedia hair > used in Theodolite
There fore correct answer is Option - 4.
Q.8. On map long form of R. F. is____
Refractive Factor.
Refractive Fraction.
Representative Fraction.
Representative Factor.
Q.9. A levelling is carried out to established the reduced level (RL) of point R
with respect to the bench mark (BM) at P. If RL of P is + 100 m, then what is the RL (m) of R? The staff reading taken are given
below.
| Staff at | BS | IS | FS | RL |
| P | 1.655 | - | - | 100.00 |
| Q | -0.950 | - | -1.500 | - |
| R | - | - | 0.75 | ? |
GATE 2014
103.355.
103.155.
101.455.
100.355
This is carried out fly levelling. There for we carry out HI at each station.
HI at station P = 100 + 1.655 = 101.655 m
RL of station Q = 101.655 + 1.500 (-1.500 staff reading means inverted staff reading)
RL of station Q = 103.155 m
Instrument is changed from this point and shifted other place. And staff reading taken with inverted staff(i.e. -0.950 )
HI at station Q = RL of Q - 0.950 = 102.205
Staff is held at point R which is below HI.
RL of station R = HI - 0.750 = 102.205 - 0.750 = 101.455m
There fore correct answer is Option - 3.
HI at station P = 100 + 1.655 = 101.655 m
RL of station Q = 101.655 + 1.500 (-1.500 staff reading means inverted staff reading)
RL of station Q = 103.155 m
Instrument is changed from this point and shifted other place. And staff reading taken with inverted staff(i.e. -0.950 )
HI at station Q = RL of Q - 0.950 = 102.205
Staff is held at point R which is below HI.
RL of station R = HI - 0.750 = 102.205 - 0.750 = 101.455m
There fore correct answer is Option - 3.
Q.10. A line PQ is measured by 30 m chain is 1230 m. At the end chain was tested and it found that chain 12 cm too short. Find correct length of line PQ.
1225.08 m.
1230 m.
1229.88 m.
none of the above
Correct Chain length = L = 30 m.
Incorrect Chain length = L' = 30 m - 0.12 m = 29.88 m.
Incorrect PQ length = 1230.
Correct PQ length = L'/L * 1230
Correct PQ length = 1225.08 m
There fore correct answer is Option - 1.
Incorrect Chain length = L' = 30 m - 0.12 m = 29.88 m.
Incorrect PQ length = 1230.
Correct PQ length = L'/L * 1230
Correct PQ length = 1225.08 m
There fore correct answer is Option - 1.
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