# Fundamental Concepts of Surveying

 Question 1
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 1 Explanation:
Reconnaissance$\rightarrow$Field observations$\rightarrow$Data analysis$\rightarrow$Map making
 Question 2
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 2 Explanation:
The principal of fixed hair tacheometry is that distances are proportional to staff intercept.
As distance increase, staff intercept also increases.
 Question 3
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 4
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 4 Explanation:
Mean sea level (MSL) is used as a reference surface for establishing the vertical control.
 Question 5
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 5 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 6
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 6 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 7
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 7 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 8
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 8 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 8 questions to complete. 