The magnetic bearing of the sun for a location at noon is . 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. $93^{\circ} 30^{\prime} \mathrm{E}$

In a survey work, three independent angle, X, Y and Z were observed with weight respectively. The weight of the sum of angles X, Y and Z is given by:

Currently no explanation available

A. $1/(\frac{1}{W_{X}}+\frac{1}{W_{Y}}+\frac{1}{W_{Z}})$

B. $(\frac{1}{W_{X}}+\frac{1}{W_{Y}}+\frac{1}{W_{Z}})$

D. $W_{X}^{2}+W_{Y}^{2}+W_{Z}^{2}$

GATE CE · Geomatics Engineering

Fundamental Concepts Of Surveying – GATE CE Geomatics EngineeringPractice Questions & PYQs

Generate GATE-level questions on Fundamental Concepts Of Surveying in Geomatics Engineering. Focus on core concepts, previous year patterns, and numerical problem-solving techniques.

14 questions · 14 PYQs · 0 AI practice · GATE CE 2027

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Q1GATE 2025NAT

A surveyor measured the distance between two points on the plan drawn to a scale of $1\ \text{cm} = 40\ \text{m}$ and the result was 468 m. Later, it was discovered that the scale used was $1\ \text{cm} = 20\ \text{m}$ . The true distance between the points (in m) is ______ (round off to the nearest integer).

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📖 Explanation

$\text{RF of wrong scale} = \frac{1}{20}$ $\text{RF of corrected scale} = \frac{1}{40}$ $\text{Measured length} = 468 , \text{m}$ $\text{Corrected length} = \left( \frac{(RF){\text{wrong scale}}}{(RF){\text{corrected scale}}} \right) \times \text{Measured length}$ $= \left( \frac{\frac{1}{20}}{\frac{1}{40}} \right) \times 468 = 936 , \text{m}$

Q2GATE 2024MCQ

What is the CORRECT match between the survey instruments/parts of instruements shown in the table and the operations carried out with them?

\begin{array}{|l|l|} \hline \text{Instruments/Parts of instruments} & \text{Operations} \\ \hline \text{P. Bubble tube} & \text{i. Tacheometry} \\ \text{Q. Plumb bob} & \text{ii. Minor movements} \\ \text{R. Tangent screw} & \text{iii. Centering} \\ \text{S. Stadia cross-wire} & \text{iv. Levelling} \\ \hline \end{array}

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Q3GATE 2023MCQ

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?

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📖 Explanation

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

Q4GATE 2023MCQ

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 ___

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📖 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 $3^{\circ} 30 \mathrm{~W}$ .

Q5GATE 2022MCQ

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

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📖 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}

Q6GATE 2022NAT

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)

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📖 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}

Q7GATE 2021MCQ

In general, the CORRECT sequence of surveying operations is

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📖 Explanation

Reconnaissance $\rightarrow$ Field observations $\rightarrow$ Data analysis $\rightarrow$ Map making

Q8GATE 2021MSQ

Which of the following is/are correct statement(s)?

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Q9GATE 2015MCQ

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:

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Q10GATE 2015MCQ

Which of the following statements is FALSE?

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Q11GATE 2014MCQ

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

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Q12GATE 2008MCQ

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

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📖 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}

Q13GATE 2008MCQ

The type of surveying in which the curvature of the earth is taken into account is called

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Q14GATE 2007MCQ

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

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📖 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}

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