Past Year Questions: Compass & Traverse Surveying

# Past Year Questions: Compass & Traverse Surveying | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE) PDF Download

Question 1: The data from a closed traverse survey PQRS (run in the clockwise direction) are given in the table

The closing error for the traverse PQRS (in degrees) is ________.    [2019 : 1 Mark, Set-II]

Solution: Assuming it as anticlockwise traverse. Mathematically sum of interior angle for a closed traverse
= (2n - 4) x 90
= (2 x 4 - 4) x 90
= 4 x 90 = 360°
Given sum of interior angles, = 88 + 92 + 94 + 89 = 363°
Then error in interior angle = 363 - 360 = 3°
Note: In this question as per clockwise traverse included angle should be taken as exterior angle. But if we take exterior angle then we get all interior angles more than 180°.

Question 2: The interior angles of four triangles are given below:

Which of the triangles are ill-conditioned and should be avoided in Triangulation surveys?    [2019 : 1 Mark, Set-I]
(a) Both Q and S
(b) Both P and S
(c) Both Q and R
(d) Both P and R
Answer: (a)
Solution: For a well conditioned triangle.
The interior angle should not be less than 30°. In this way, triangle ‘Q’ & S having less angles (acute angle).
⇒ Q & S are ill-conditioned.

Question 3: The following details refer to a closed traverse:

The length and direction (whole circle bearing) of closure, respectively are    [2018 : 2 Marks, Set-I]
(a) 1 m and 90°
(b) 2 m and 90°
(c) 1 m and 270°
(d) 2 m and 270°
Answer: (a)
Solution:
∑L = W northing- E southing
= (101 + 419) - (437 + 83)
= 0
∑D = E Easting - E Westing
= (173 + 558) - (96 + 634)
= 1 m
∴ Length of closure

And, direction of closure,

Hence, θ lies in I quadrant and θ = 90°

Question 4: The observed bearings of a traverse are given below

The station(s) most likely to be affected by the local attraction is/are     [2017 : 2 Marks, Set-I]
(a) Only R
(b) Only S
(c) Rand S
(d) Pand Q
Answer: (a)
Solution:

Station R is most likely to be affected.
Since difference of FB and BB for PQ and ST is 180°.
Hence, P, Q, S, T are free from local attraction.

Question 5: The reduced bearing of a 10 m long line is N30°E. The departure of the line is    [2016 : 1 Mark, Set-II]
(a) 10.00 m
(b) 8.66 m
(c) 7.52 m
(d) 5.00 m
Answer: (d)
Solution:

The departure of the line,
D = l sin θ
= 10 sin 30°
= 10/2 = 5m

Question 6: The bearings of two inaccessible stations, S1 (Easting 500 m, Northing 500 m) and S2 (Easting 600 m, Northing 450 m) from a station S3 were observed as 225° and 153° 26' respectively. The independent Easting (in m) of station S3 is:    [2015 : 2 Marks, Set-II]
(a) 450.000
(b) 570.710
(c) 550.000
(d) 650.000
Answer: (c)
Solution:

Let S1S3 = l1 and S2S3 = l2
Northing of S3:
500 + l1cos45°= 450 + l2cos26°34'
l1 cos 45° - l2cos 26° 34' = - 50    ..(i)
Easting of S3
500 + l1 sin 45° = 600 - l2sin26°34'
l1sin45° + l2sin26°34' = 100     ..(ii)
Solving eq. (i) and eq. (ii),
l= 70.707
l2 = 111.802
∴ Easting of S3 = 500 + 70.707 sin45°
= 550m

Question 7: In a region with magnetic declination of 2°E, the magnetic Fore Bearing (FB) of a line AB was measured as N79°50'E. There was local attraction at A. To determine the correct magnetic bearing of the line, a point 0 was selected at which there was no local attraction. The magnetic FB of line AO and OA were observed to be S52°40'E and N50°20'W, respectively. What is the true FB of line AB?    [2015 : 2 Marks, Set-I]
(a) N81°50'E
(b) N82°10'E
(c) N84°10'E
(d) N77°50'E
Answer: (c)
Solution: Magnetic Declination,δ = 2°E
Magnetic FB of AB = N 79°50' E
To find local attraction at station A
As station O is free from local attraction
Hence FB of OA will be correct,
Correct FB of OA = N 50°20' W ≌ 309°40'
∴ Correct BB of OA = 129°40'
∵ Obserbed FB of AO
= Observed BB of OA
= S52°40'E
= 127°20
Error = Observed bearing - Correct bearing
= 127°20' - 129°40'
= 2°20'
Correction = +2°20'
Local attraction at station A
= +2°20' ≌ 2°20' E
∴ Magnetic FB of AB = N 79°50'E
δ = 2°E and local attraction = 2°20'E
∴ TB of FB of AB = 79°50' + 2°20' + 2°
= N84°10'E

Question 8: In a closed loop traverse of 1 km total length, the closing errors in departure and latitude are 0.3 m and 0.4 m, respectively. The relative precision of this traverse will be    [2015 : 1 Mark, Set-I]
(a) 1 : 5000
(b) 1 : 4000
(c) 1 : 3000
(d) 1 : 2000
Answer: (d)
Solution: Relative precision
where, p = Perimeter of traverse
e = Closing error

Question 9: Group-I lists tool/instrument while Group-ll lists the method of surveying. Match the tool/ instrument with the corresponding method of surveying.    [2014 : 2 Marks, Set-I]

Answer: (d)
Solution: 1. Alllade: It is a straight edge ruler used in plane table surveying whichis used for sighting the objects and drawing the line son the drawing sheet.
2. Arrow: Arrows are used to mark the position of end of the hain or tape on the ground.
3. Bubble tube: It is used to check the level of the instrument. When the bubble of the tube comes in the centre, then the instrument in levelled.
4. Stadia hair: Lines on diaphragm, the intercept ‘S’ between 2 stadia hairs of a vertically held rod gives the distance between tacheometer and the road as
D = kS (k = stadia constant)

Question 10: A Theodolite is placed at A and a 3 m long vertical staff is held at B. The depression angle made at reading of 2.5 m marking on staff is 6°10'. The horizontal distance between A and B is 2200 m. The height of instrument at A is 1.1 m and reduced level of point A is 880.88 m using curvature correction and refraction correction determine the R.L. of point B(in m)?    [2013 : 2 Marks]
Solution: RL of A = 880.88 m
RL of plane of collimation
= 880.88 + 1.1 = 881.98 m

Here, D = Distance between theodolite and staff in km
∴ RL of B = 881.98-237.701
- (2.5 x 0.0673 x 2.22)
= 642.105 m

Question 11: Bearing of the given system is shown below:

Applying correction due to local attraction, the correct bearing of line BC will be    [2013 : 2 Marks]
(a) 48°15'
(b) 50° 15'
(C) 49° 15'
(d) 48°45'
Answer: (d)
Solution: The difference between back bearing and fore bearing of line DE
= 258° 30' - 78° 30' = 180°
∴ Station D and E are free from local attraction
Fore bearing of line EA = 216° 30'
∴ Correct back bearing of line EA
= 216° 30' - 180° = 36° 30'
∴ Error at A = 31° 45' - 36° 30' = -4°45'

Question 12: The latitude and departure of a line AB are +78 m and -45.1 m respectively. The whole circle bearing of the line AB is    [2013 : 1 Mark]
(a) 30°
(b) 150°
(c) 210°
(d) 330°
Answer: (d)
Solution:

Since the latitude of line is positive and departure is negative, the line lies in the fourth quadrant.
∴ L*cosθ = 78
L*sinθ = -45.1
⇒ tanθ = -0.578
θ = -30°
∴ WCB of AB = 360° - 30°
= 330°

Question 13: The observations from a closed loop traverse around an obstacle are

What is the value of the missing measurement (rounded off to the nearest 10 mm)?    [2011 : 2 Marks]
(a) 396.86 m
(b) 396.79 m
(c) 396.05 m
(d) 396.94 m

Answer: (b)
Solution: In a closed loop travers, the algebraic sum of all the latitudes should be equal to zeroi.e., ∑L = 0
⇒ Lcos 33.7500°+ 300 cos 86.3847° + 354.524 cos 169.3819° + 450 cos 243.900° + 268 cos 317.5° = 0
⇒ Lcos 33.75° = 329.9166
⇒ L = 396.79 m

The document Past Year Questions: Compass & Traverse Surveying | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE) is a part of the Civil Engineering (CE) Course Topic wise GATE Past Year Papers for Civil Engineering.
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## FAQs on Past Year Questions: Compass & Traverse Surveying - Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE)

 1. What is compass surveying?
Ans. Compass surveying is a method of surveying where a magnetic compass is used to determine the direction of lines, angles, and bearings on the land. It is commonly used for small-scale surveys and is based on the principle of magnetic needle alignment with the Earth's magnetic field.
 2. What is traverse surveying?
Ans. Traverse surveying is a method of surveying that involves a series of connected survey lines or measurements made between established control points. It is used to determine the coordinates, angles, and distances of various points on the land. Traverse surveying is commonly used for large-scale surveys and is more accurate compared to compass surveying.
 3. What are the advantages of compass surveying?
Ans. Compass surveying has several advantages, such as its simplicity, portability, and low cost. It does not require complex equipment and can be easily carried out in the field. Additionally, compass surveying is suitable for rough terrains and areas with limited access, making it a practical choice for certain surveying tasks.
 4. What are the limitations of traverse surveying?
Ans. Traverse surveying has a few limitations that should be considered. One limitation is the accumulation of errors, as any mistakes made in the initial measurements can affect the entire survey. Additionally, traverse surveying is time-consuming and requires careful calculations to ensure accuracy. It is also dependent on good weather conditions for accurate measurements.
 5. How is a compass traverse survey different from a chain traverse survey?
Ans. A compass traverse survey uses a magnetic compass to determine the direction of lines and angles, while a chain traverse survey relies on measuring distances using a chain or tape. The compass traverse is more suitable for rough terrains and areas with limited access, while the chain traverse is more accurate and precise in determining distances. Both methods have their own advantages and are chosen based on the specific requirements of the survey.

## Topic wise GATE Past Year Papers for Civil Engineering

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