Geomatics Engineering

Question 1
The error in measuring the radius of a 5 cm circular rod was 0.2%. If the cross-sectional area of the rod was calculated using this measurement, then the resulting absolute percentage error in the computed area is______. (round off to two decimal places)
A
0.25
B
0.40
C
0.67
D
0.83
GATE CE 2022 SET-2      Fundamental Concepts of Surveying
Question 1 Explanation: 
\begin{aligned} r&=5 \\ e_r&=\frac{0.2}{100} \times 5=0.01cm \\ A&=\pi r^2 \\ e_A&=2 \pi r.e_r \end{aligned}
Absolute perecentage error in computed area
\begin{aligned} &=\frac{e_A}{A} \times 100 \\ &=\frac{2 \pi r.e_r}{\pi r^2} \times 100\\ &=2 \times \left ( \frac{e_r}{r} \times 100 \right )\\ &=2 \times 0.2=0.4 \end{aligned}
Question 2
If the magnetic bearing of the Sun at a place at noon is S2^{\circ}E, the magnetic declination (in degrees) at that place is
A
2^{\circ}E
B
2^{\circ}W
C
4^{\circ}E
D
4^{\circ}W
GATE CE 2022 SET-2      Theodolites, Compass and Traverse Surveying
Question 2 Explanation: 
MB=S2^{\circ}E=180^{\circ}-2^{\circ}=178^{\circ}
TB= 180^{\circ}
Declination, \delta =TB-MB=180-178=2^{\circ} \; or \; 2^{\circ}E
Question 3
The bearing of a survey line is N31^{\circ}17'W. Its azimuth observed from north is ______ deg. (round off to two decimal places)
A
328.71
B
458.25
C
124.65
D
625.25
GATE CE 2022 SET-1      Theodolites, Compass and Traverse Surveying
Question 3 Explanation: 


WCB=360-\left ( 31+\frac{17}{60} \right )^{\circ}=329.716^{\circ}
Question 4
A line between stations P and Q laid on a slope of 1 in 5 was measured as 350 m using a 50 m tape. The tape is known to be short by 0.1 m.
The corrected horizontal length (in m) of the line PQ will be
A
342.52
B
349.3
C
356.2
D
350.7
GATE CE 2022 SET-1      Fundamental Concepts of Surveying
Question 4 Explanation: 


Horizontal distance of line
PQ=350 \cos \theta =\frac{350 \times 5}{\sqrt{26}}=343.20m
Tape is 0.1 m short.
Nominal length of tape, l = 50 m
Actual length of tape, l' = 50 - 0.1 = 49.9 m
Corrected horizontal length of line PQ
\begin{aligned} &=\left ( \frac{l'}{l} \right ) \times 343.2\\ &=\left ( \frac{49.9}{50} \right )\times 343.2\\ &=342.517\simeq 342.52m \end{aligned}
Question 5
An aerial photograph is taken from a flight at a height of 3.5 km above mean sea level, using a camera of focal length 152 mm. If the average ground elevation is 460 m above mean sea level, then the scale of the photograph is
A
1 : 20000
B
01:20
C
1 : 100000
D
1.986111111
GATE CE 2022 SET-1      Remote Sensing, GIS, GPS and Photogrammetry
Question 5 Explanation: 
\begin{aligned} H &=3.5Km=3500m\\ f&=152mm \\ h_{avg}&=460m \\ Scale&=\frac{f}{H-h_{avg}} \\ &= \frac{152 \times 10^{-3}}{3500-460}\\ &=\frac{1}{20000} \end{aligned}
Question 6
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      Theodolites, Compass and Traverse Surveying
Question 6 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 7
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      Theodolites, Compass and Traverse Surveying
Question 7 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 8
In general, the CORRECT sequence of surveying operations is
A
Field observations\rightarrow Reconnaissance\rightarrow Data analysis\rightarrow Map making
B
Data analysis\rightarrow Reconnaissance\rightarrow Field observations \rightarrow Map making
C
Reconnaissance\rightarrow Field observations \rightarrow Data analysis \rightarrow Map making
D
Reconnaissance\rightarrow Data analysis \rightarrow Field observations \rightarrow Map making
GATE CE 2021 SET-2      Fundamental Concepts of Surveying
Question 8 Explanation: 
Reconnaissance\rightarrowField observations\rightarrowData analysis\rightarrowMap making
Question 9
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      Theodolites, Compass and Traverse Surveying
Question 9 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 10
Which of the following is/are correct statement(s)?
A
Back Bearing of a line is equal to Fore Bearing \pm 180^{\circ}
B
If the whole circle bearing of a line is 270^{\circ}, its reduced bearing is 90^{\circ} \mathrm{NW}
C
The boundary of water of a calm water pond will represent contour line
D
In the case of fixed hair stadia tachometry, the staff intercept will be larger, when the staff is held nearer to the observation point
GATE CE 2021 SET-1      Fundamental Concepts of Surveying
Question 10 Explanation: 
The principal of fixed hair tacheometry is that distances are proportional to staff intercept.
As distance increase, staff intercept also increases.
There are 10 questions to complete.

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