# Geomatics Engineering

 Question 1
A delivery agent is at a location R. To deliver the order, she is instructed to travel to location $P$ along straight-line paths of $\mathrm{RC}, \mathrm{CA}, \mathrm{AB}$ and $\mathrm{BP}$ of $5 \mathrm{~km}$ each. The direction of each path is given in the table below as whole circle bearings. Assume that the latitude (L) and departure (D) of $R$ is $(0,0) \mathrm{km}$. What is the latitude and departure of $P$ (in km, rounded off to one decimal place)?
$\begin{array}{|c|c|c|c|c|} \hline Paths & RC & CA & AB & BP \\ \hline \begin{array}{c}\text {Directions} \\ \text {(in degrees)}\end{array} & 120 & 0 & 90 & 240 \\ \hline \end{array}$
 A $L=2.5 ; D=5.0$ B $L=0.0 ; D=5.0$ C $L=5.0 ; D=2.5$ D $L=0.0 ; D=0.0$
GATE CE 2023 SET-2      Theodolites, Compass and Traverse Surveying
Question 1 Explanation: Latitude of ' $C$ ' $=-5 \sin 30^{\circ}=-2.5 \mathrm{Km}$
Departure of ' $C$ ' $=5 \cos 30^{\circ}=+4.33 \mathrm{Km}$
Latitude of ' $\mathrm{B}$ ' $=$ Latitude of ' $\mathrm{C}$ ' $+5 \mathrm{Km}$ $=-2.5+5=+2.5 \mathrm{Km}$
Departure of 'B' $=$ Departure of ' $C$ ' $+5 \mathrm{Km}$ $=+4.33+5=+9.33 \mathrm{Km}$
Latitude of ' $P$ ' = Latitude of ' $B$ ' $-5 \sin 30^{\circ}$ $=+2.5-2.5=0$
Departure of ' $P$ ' $=$ Departure of ' $B$ ' $-5 \sin 30^{\circ}$ $=9.33-4.33=+5 \mathrm{Km}$
$\therefore \quad \mathrm{L}=0.0, \quad \mathrm{D}=5.0$
 Question 2
If the size of the ground area is $6 \mathrm{~km} \times 3 \mathrm{~km}$ and the corresponding photo size in the aerial photograph is $30 \mathrm{~cm} \times 15 \mathrm{~cm}$, then the scale of the photograph is 1 : ____ (in integer).
 A 15000 B 20000 C 25000 D 30000
GATE CE 2023 SET-2      Remote Sensing, GIS, GPS and Photogrammetry
Question 2 Explanation:
Scale of Photograph $(\mathrm{S})=\frac{\text { Photo distance }}{\text { Ground distance }}$

For Length scale :
\begin{aligned} S & =\frac{30 \mathrm{~cm}}{6 \mathrm{~cm}} \\ & =\frac{30 \mathrm{~cm}}{60 \times 10^{4} \mathrm{~cm}} \end{aligned}
$\Rightarrow \quad \mathrm{S}=\frac{1}{20000}$

\begin{aligned} S & =\frac{15 \mathrm{~cm}}{3 \mathrm{Km}} \\ & =\frac{15 \mathrm{~cm}}{30 \times 10^{4} \mathrm{~cm}} \end{aligned}
$S=\frac{1}{20000}$
$\therefore$ Scale of photograph, $S=1: 20000$

 Question 3
The magnetic bearing of the sun for a location at noon is $183^{\circ} 30^{\prime}$. If the sun is exactly on the geographic meridian at noon, the magnetic declination of the location is ___
 A $3^{\circ} 30^{\prime} \mathrm{W}$ B $3^{\circ} 30^{\prime} \mathrm{E}$ C $93^{\circ} 30^{\prime} \mathrm{W}$ D $93^{\circ} 30^{\prime} \mathrm{E}$
GATE CE 2023 SET-2      Fundamental Concepts of Surveying
Question 3 Explanation:
At any place 12 O'clock sun will be exactly over the true meridian of that place. Declination $=183^{\circ} 30^{\prime}-180^{\circ}=3^{\circ} 30^{\prime} \mathrm{W}$
Magnetic declination will be $3^{\circ} 30 \mathrm{~W}$.
 Question 4
Trigonometric levelling was carried out from two stations $P$ and $Q$ to find the reduced level (R.L.) of the top of hillock, as shown in the table. The distance between stations $P$ and $Q$ is $55 \mathrm{~m}$. Assume stations $P$ and $Q$, and the hillock are in the same vertical plane. The R. L. of the top of the hillock (in$\mathrm{m}$ ) is _____ (round off to three decimal places)
$\begin{array}{|l|l|l|l|} \hline Station &\begin{array}{c}\text{Vertical} \\ \text{angle of} \\ \text {the top of} \\ \text{hillock }\end{array} &\begin{array}{c}\text {Staff} \\ \text{reading} \\ \text {on} \\ \text{benchmark}\end{array} &\begin{array}{c}\text {R.L. of} \\ \text {benchmark}\end{array} \\ \hline P &18^{\circ} 45^{\prime} &2.340 \mathrm{~m} &100.000 \mathrm{~m} \\ \hline Q &12^{\circ} 45^{\prime} &1.660 \mathrm{~m}& \\ \hline \end{array}$
 A 124.365 B 137.682 C 253.142 D 254.325
GATE CE 2023 SET-1      Levelling and Contouring
Question 4 Explanation: While taking reading from $P$,
RL of hillock $=H I_P+x \tan 18^{\circ} 45^{\prime}$
$=100+2.34+x \tan 18^{\circ} 45^{\prime}$

While taking reading from $Q$.
$\mathrm{RL}$ of hillock $=\mathrm{HI}_Q+(\mathrm{x}+55) \tan 12^{\circ} 45^{\prime}$
$=100+1.66+(x +55)\tan 12^{\circ} 45^{\prime}$
Equating both,
$100+2.34+x \tan 18^{\circ} 45^{\prime}=100+1.66+(x +55)\tan 12^{\circ} 45^{\prime}$
$2.34+0.339 x=1.66+12.445+0.226 x$
$\therefore x=104.115$

$\quad \therefore \quad$ RL of hillock
$=100+2.34+104.115 \tan 18^{\circ} 45^{\prime}$
$=137.682$
 Question 5
A student is scanning his 10 inch $\times$ 10 inch certificate at 600 dots per inch (dpi) to convert it to raster. What is the percentage reduction in number of pixels if the same certificate is scanned at 300 dpi?
 A 62 B 88 C 75 D 50
GATE CE 2023 SET-1      Fundamental Concepts of Surveying
Question 5 Explanation:
DPI = Dots Per Inch, it should be noted that DPI is not dots per square inch.
$\%$ reduction in number of pixels
$=\frac{10 \times 600 \times 10 \times 600-100 \times 300 \times 10 \times 300}{10 \times 600 \times 10 \times 600} \times 100$
$=\frac{600^{2}-300^{2}}{600^{2}} \times 100$
$=75 \%$

There are 5 questions to complete.

### 3 thoughts on “Geomatics Engineering”

1. Thankyou very much for providing Pyq s with neat explanation…

2. 3. 