Theodolites, Compass and Traverse Surveying

Question 1
For a given traverse, latitudes and departures are calculated and it is found that sum of latitudes is equal to +2.1 m and the sum of departures is equal to -2.8 m. The length and bearing of the closing error, respectively, are
A
3.50 \mathrm{~m} \text { and } 53^{\circ} 7^{\prime} 48^{\prime \prime} \mathrm{NW}
B
2.45 \mathrm{~m} \text { and } 53^{\circ} 7^{\prime} 48^{\prime \prime} \text { NW }
C
0.35 \mathrm{~m} \text { and } 53.13^{\circ} \mathrm{SE}
D
3.50 \mathrm{~m} \text { and } 53.13^{\circ} \mathrm{SE}
GATE CE 2021 SET-2   Geometics Engineering
Question 1 Explanation: 
\begin{aligned} e_{L} &=+2.1 \mathrm{~m} \\ e_{D} &=-2.8 \mathrm{~m} \\ e &=\sqrt{e_{L}^{2}+e_{D}^{2}} \\ &=\sqrt{(2.1)^{2}+(2.8)^{2}}=3.5 \mathrm{~m} \\ \text { Bearing of closing error } &=\tan ^{-1}\left(\frac{e_{D}}{e_{L}}\right) \\ &=\tan ^{-1}\left(\frac{-2.8}{2.1}\right)=-53.13^{\circ} \\ &=53^{\circ} 7^{\prime} 48^{\prime \prime} \mathrm{NW} \end{aligned}
Question 2
A horizontal angle \theta is measured by four different surveyors multiple times and the values reported are given below.
\begin{array}{|c|c|c|} \hline \text { Surveyor } & \text { Angle } \theta & \text { Number of observations } \\ \hline 1 & 36^{\circ} 30^{\prime} & 4 \\ \hline 2 & 36^{\circ} 00^{\prime} & 3 \\ \hline 3 & 35^{\circ} 30^{\prime} & 8 \\ \hline 4 & 36^{\circ} 30^{\prime} & 4 \\ \hline \end{array}
he most probable value of the angle \theta ( in degree, round off to two decimal placesis ________
A
12
B
28
C
36
D
44
GATE CE 2021 SET-2   Geometics Engineering
Question 2 Explanation: 
\begin{aligned} \mathrm{MPV} &=\frac{\left(36^{\circ} 30^{\prime} \times 4\right)+\left(36^{\circ} \times 3\right)+\left(35^{\circ} 30^{\prime} \times 8\right)+\left(36^{\circ} 30^{\prime} \times 4\right)}{4+3+8+4} \\ &=36^{\circ} \end{aligned}
Question 3
Traversing is carried out for a closed traverse PQRS. The internal angles at vertices P, Q, R and S are measured as 92^{\circ} ,62^{\circ} ,123^{\circ} and 77^{\circ} , respectively. If fore bearing of line PQ is 27^{\circ} , fore bearing of line RS (in degrees, in integer) is _________
A
258
B
753
C
159
D
218
GATE CE 2021 SET-1   Geometics Engineering
Question 3 Explanation: 


\begin{array}{l} Q=\left[\begin{array}{l} B B \text { of } P Q=27^{\circ}+180^{\circ}=207^{\circ} \\ F B \text { of } Q R=207^{\circ}-68^{\circ}=139^{\circ} \end{array}\right. \\ R=\left[\begin{array}{l} B B \text { of } Q R=139^{\circ}+180^{\circ}=319^{\circ} \\ F B \text { of } R S=319^{\circ}-123^{\circ}=196^{\circ} \end{array}\right. \end{array}

\begin{array}{l} Q=\left[\begin{array}{l} B B \text { of } P Q=27^{\circ}+180^{\circ}=207^{\circ} \\ F B \text { of } Q R=207^{\circ}+68^{\circ}=275^{\circ} \end{array}\right. \\ R=\left[\begin{array}{l} B B \text { of } Q R=275^{\circ}-180^{\circ}=95^{\circ} \\ F B \text { of } R S=95^{\circ}+123^{\circ}=218^{\circ} \end{array}\right. \end{array}
Question 4
A theodolite is set up at station A. The RL of instrument axis is 212.250 m. The angle of elevation to the top of a 4 m long staff, held vertical at station B, is 7^{\circ}. The horizontal distance between station A and B is 400 m. Neglecting the errors due to curvature of earth and refraction, the RL (in m, round off to three decimal places) of station B is __________
A
257.363
B
145.126
C
472.156
D
324.422
GATE CE 2020 SET-2   Geomatics Engineering
Question 4 Explanation: 


\begin{aligned} V&=400 \tan 7^{\circ}\\ &=49.113\\ x&=(49.113-4)=45.113\\ RL_B&=212.25+45.113\\ &=257.363m \end{aligned}
Question 5
A theodolite was set up at a station P. The angle of depression to a vane 2 m above the foot of a staff held at another station Q was 45^{\circ}. The horizontal distance between stations P and Q is 20 m. The staff reading at a benchmark S of RL 433.050 m is 2.905 m. Neglecting the errors due to curvature and refraction, the RL of the station Q (in m), is
A
413.05
B
413.955
C
431.05
D
435.955
GATE CE 2020 SET-2   Geomatics Engineering
Question 5 Explanation: 


\begin{aligned} \frac{x}{20}&=\tan 45^{\circ}\\ x&=20m\\ RL \; of \; Q&=433.05+2.905-x-2\\ &=433.05+2.905-20-2\\ &=413.955m \end{aligned}
Question 6
The length and bearings of a traverse PQRS are:

The length of line segment SP (in m, round off to two decimal places), is ________.
A
33.07
B
25.36
C
47.78
D
44.79
GATE CE 2020 SET-1   Geomatics Engineering
Question 6 Explanation: 
\begin{aligned} \Delta L&= 40 \cos 80^{\circ}+50 \cos 10^{\circ}+30 \cos 210^{\circ}\\ &=30.20 \\ \Delta D&=40 \sin 80^{\circ}+50 \sin 10^{\circ}+30 \sin 210^{\circ} \\ &=33.07 \\ \text{Length, \; SP}&=\sqrt{\Delta L^2+\Delta D^2}\\ &=44.79m \end{aligned}
Question 7
An open traverse PQRST is surveyed using theodolite and the consecutive coordinates obtained are given in the table

If the independent coordinates (Northing, Easting) of station P are (400 m, 200 m) the independent coordinates (in m) of station T, are
A
194.7, 370.1
B
205.3, 429.9
C
394.7, 170.1
D
405.3, 229.9
GATE CE 2020 SET-1   Geomatics Engineering
Question 7 Explanation: 


\begin{aligned} \Delta L &=-5.3 \\ \Delta D&=-29.9 \\ T, \; \text{Northing}&\{400+(-5.3)\}=394.7 \\ T, \; \text{Easting}&\{200+(-29.9)\}=170.1\\ &T[394.7m, 170.1m] \end{aligned}
Question 8
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 ___
A
2
B
3
C
4
D
5
GATE CE 2019 SET-2   Geomatics Engineering
Question 8 Explanation: 
Assuming it as anticlockwise traverse.
Mathematically sum of interior angle for a closed traverse =(2n-4)\times 90 = (2 \times 4 -4) \times 90 = 360^{\circ}
Given sum of interior angles, = 88 + 92 + 94 + 89 = 363^{\circ}
Then error in interior angle = 363 - 360 = 3^{\circ}
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^{\circ}.
Question 9
The interior angles of four triangles are given below:

Which of the triangles are ill-conditioned and should be avoided in Triangulation surveys?
A
Both P and R
B
Both Q and R
C
Both P and S
D
Both Q and S
GATE CE 2019 SET-1   Geomatics Engineering
Question 9 Explanation: 
For an ill conditioned traingle in triangulation survey, any angle can be less than 38^{\circ}, and can be greater than 38^{\circ}.
For traingles Q and S, the above condition is valid.
Question 10
The following details refer to a closed traverse:

The length and direction (whole circle bearing) of closure, respectively are
A
1 m and 90^{\circ}
B
2 m and 90^{\circ}
C
1 m and 270^{\circ}
D
2 m and 270^{\circ}
GATE CE 2018 SET-1   Geomatics Engineering
Question 10 Explanation: 
\begin{aligned} \text { Latitude }_{P Q} &=L_{P Q}=-437 \mathrm{m} \\ L_{S P} &=-83 \mathrm{m} \\ L_{O R} &=+101 \mathrm{m}\\ L_{R S} &=+419 \mathrm{m} \\ \Sigma \text { Latitudes } &=-437-83+101+419 \\ &=0 \mathrm{m} \\ \text { Departure }_{P O} &=D_{P O}=+173 \mathrm{m} \\ D_{O R} &=+558 \mathrm{m} \\ D_{8 S} &=-96 \mathrm{m} \\ D_{S P} &=-634 \mathrm{m} \end{aligned}
\Sigma Departures =1 m For closure of traverse
\Sigma Latitude =0
\Sigma Departure =0
Then,
Departure of closure =-1 m,
Latitude of closure =0


So the length and direction (whole circle bearing of closure is 1m and 270^{\circ} respectively.
There are 10 questions to complete.

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