# Fundamental Concepts of Surveying

 Question 1
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   Geomatics Engineering
Question 1 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 2
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   Geomatics Engineering
Question 2 Explanation:
DPI = Dots Per Inch, it should be noted that DPI is not dots per square inch.
$\%$ reduction in number of pixels
$=\frac{10 \times 600 \times 10 \times 600-100 \times 300 \times 10 \times 300}{10 \times 600 \times 10 \times 600} \times 100$
$=\frac{600^{2}-300^{2}}{600^{2}} \times 100$
$=75 \%$

 Question 3
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 0.4 C 0.67 D 0.83
GATE CE 2022 SET-2   Geometics Engineering
Question 3 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 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   Geometics Engineering
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
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 5 Explanation:
Reconnaissance$\rightarrow$Field observations$\rightarrow$Data analysis$\rightarrow$Map making

There are 5 questions to complete.