Fundamental Concepts of Surveying

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   Geometics Engineering
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
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   Geometics Engineering
Question 2 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 3
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   Geometics Engineering
Question 3 Explanation: 
Reconnaissance\rightarrowField observations\rightarrowData analysis\rightarrowMap making
Question 4
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   Geometics Engineering
Question 4 Explanation: 
The principal of fixed hair tacheometry is that distances are proportional to staff intercept.
As distance increase, staff intercept also increases.
Question 5
In a survey work, three independent angle, X, Y and Z were observed with weight W_{X},W_{Y},W_{Z} respectively. The weight of the sum of angles X, Y and Z is given by:
A
1/(\frac{1}{W_{X}}+\frac{1}{W_{Y}}+\frac{1}{W_{Z}})
B
(\frac{1}{W_{X}}+\frac{1}{W_{Y}}+\frac{1}{W_{Z}})
C
W_{X}+W_{Y}+W_{Z}
D
W_{X}^{2}+W_{Y}^{2}+W_{Z}^{2}
GATE CE 2015 SET-1   Geomatics Engineering
Question 6
Which of the following statements is FALSE?
A
Plumb line is along the direction of gravity
B
Mean Sea Level (MSL) is used as a reference surface for establishing the horizontal control
C
Mean Sea Level (MSL) is a simplification of Geoid
D
Geoid is an equi-potential surface of gravity
GATE CE 2015 SET-1   Geomatics Engineering
Question 6 Explanation: 
Mean sea level (MSL) is used as a reference surface for establishing the vertical control.
Question 7
The survey carried out to delineate natural features,such as hills, rivers, forests and manmade features, suchas towns, villages, buildings, roads, transmission lines and canals is classified as
A
engineering survey
B
geological survey
C
land survey
D
topographic survey
GATE CE 2014 SET-2   Geomatics Engineering
Question 7 Explanation: 
Tropographic survey is done to determine the natural features of a country such as rivers, strains, lakes, wood, hills etc. and artificial features such as roads, railways, canals, towns and villages.
Question 8
The plan of a survey plotted to a scale of 10 m to 1 cm is reduced in such a way that a line originally 10 cm long now measures 9 cm. The area of the reduced plan is measured as 81 cm^{2}. The actual area (m^{2}) of the survey is
A
10000
B
6561
C
1000
D
656
GATE CE 2008   Geomatics Engineering
Question 8 Explanation: 
Shrinking factor = \frac{9}{10}=0.9
Reduced plan area =(\text { Shrinkage factor })^{2}
\times Actual plan area
\Rightarrow \quad 81=(0.9)^{2} \times Actual plan area
\Rightarrow Actual plan area =100 \mathrm{cm}^{2}
\therefore \quad
Actual area of survey in \mathrm{m}^{2}
\begin{aligned} &=100 \times(10)^{2} \\ &=10000 \end{aligned}
Question 9
The type of surveying in which the curvature of the earth is taken into account is called
A
Geodetic surveying
B
Plane surveying
C
Preliminary surveying
D
Topographical surveying
GATE CE 2008   Geomatics Engineering
Question 9 Explanation: 
Geodetic surveying is that type of surveying in which the shape of the earth is taken into account. Geodetic surveying involves spherical trigonometry.
Question 10
The plan of a map was photo copied to a reduced size such that a line originally 100 mm, measures 90 mm. 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
GATE CE 2007   Geomatics Engineering
Question 10 Explanation: 
\begin{aligned} \text { Reduction factor }&=\frac{90}{100}=0.9 \\ \text { Revised scale }&=\text { Original scale }\\ &\times \text { Reduction factor }\\ =\frac{1}{1000} \times 0.9&=\frac{1}{1000 / 0.9}=\frac{1}{1111} \end{aligned}
There are 10 questions to complete.

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