Q21GATE 2018MSQ

The following details refer to a closed traverse: The length and direction (whole circle bearing) of closure, respectively are

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📖 Explanation

\begin{aligned} \text { Latitude }_{P Q} &=L_{P Q}=-437 \mathrm{m} \\ L_{S P} &=-83 \mathrm{m} \\ L_{O R} &=+101 \mathrm{m}\\ L_{R S} &=+419 \mathrm{m} \\ \Sigma \text { Latitudes } &=-437-83+101+419 \\ &=0 \mathrm{m} \\ \text { Departure }_{P O} &=D_{P O}=+173 \mathrm{m} \\ D_{O R} &=+558 \mathrm{m} \\ D_{8 S} &=-96 \mathrm{m} \\ D_{S P} &=-634 \mathrm{m} \end{aligned}

$\Sigma$ Departures =1 m For closure of traverse $\Sigma$ Latitude =0 $\Sigma$ Departure =0 Then, Departure of closure =-1 m, Latitude of closure =0 So the length and direction (whole circle bearing of closure is 1m and $270^{\circ}$ respectively.

Q23GATE 2017MCQ

The method of orientation used, when the plane table occupies a position not yet located on the map, is called as

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Q24GATE 2017MCQ

The observed bearings of a traverse are given below: The stations(s) most likely to be affected by the local attraction is/are

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Q25GATE 2016MCQ

The reduced bearing of a 10 m long line is $N30^{\circ}E$ . The departure of the line is

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📖 Explanation

The departure of the line,

\begin{aligned} D &=l \sin \theta \\ &=10 \sin 30^{\circ} \\ &=\frac{10}{2}=5 \mathrm{m} \end{aligned}

Q26GATE 2015MCQ

In a region with magnetic declination of $2^{\circ}E$ , the magnetic Force Bearing (FB) of a line AB was measured as $N79^{\circ}{50}'E$ . There was local attraction at A. To determine the correct magnetic bearing to the line, a point O was selected at which there was no local attraction. The magnetic FB of line AO and OA were observed to be $S 52^{\circ}{40}'E$ and $N 50^{\circ}{20}'W$ respectively. What is the true FB of line AB?

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📖 Explanation

Magnetic Declination, $\delta=2^{\circ} \mathrm{E}$ Magnetic FB of AB $=\mathrm{N} 79^{\circ} 50'E$ To find local atraction at station A As station O is free from local attraction Heance FB of OA will be correct Correct FB of OA $= N 50^{\circ}20^{\circ}W=309^{\circ}40^{\circ}$ $\therefore$ Correct BB of $OA = 120^{\circ}40^{\circ}$ $\because$ Obserbed FB of AO = Observed BB of OA $=120^{\circ}40^{\circ}$ $127^{\circ}20^{\circ}$ Error $= MB = TB =127^{\circ}20^{\circ}=120^{\circ}40^{\circ}$ $==2^{\circ}20^{\circ}$ Correction $=+2^{\circ}20^{\circ}$ Local attraction at station A $=+2^{\circ}20^{\circ}=2^{\circ}20^{\circ}E$ $\therefore$ Magnetic FB of $AB=N76^{\circ}50^{\circ}E$ $\delta = 2^{\circ}E$ and local attraction $=2^{\circ}20^{\circ}E$ $\therefore$ TB of FB of AB $= 70^{\circ}50^{\circ}+2^{\circ}20^{\circ}+2^{\circ}$ $=N84^{\circ}10^{\prime}E$

Q27GATE 2015MCQ

The bearings of two inaccessible stations, $S_{1}$ (Easting 500 m, Northing 500 m) and $S_{2}$ (Easting 600 m, Northing450 m) from a station $S_{3}$ were observed as $225^{\circ}$ and $153^{\circ}{26}'$ respectively. The independent Easting (in m) of station $S_{3}$ is:

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📖 Explanation

$Let S_{1} S_{3}=l_{1}\text{ and }S_{2} S_{3}=l_{2}$ Northing of $\mathrm{S}_{3}:$

\begin{aligned} &500+l_{1} \cos 45^{\circ}=450+l_{2} \cos 26^{\circ} 34^{\prime} \\ &l_{1} \cos 45^{\circ}-l_{2} \cos 26^{\circ} 34^{\prime}=-50 &\ldots(i) \end{aligned}

Easting of $\mathrm{S}_{3}:$

\begin{aligned} 500+l_{1} \sin 45^{\circ}=600-l_{2} \sin 26^{\circ} 34^{\prime} \\ l_{1} \sin 45^{\circ}+l_{2} \sin 26^{\circ} 34^{\prime}=100 &\ldots(ii) \end{aligned}

Solving eq. (i) and eq. (ii),

\begin{aligned} l_{1}&=70.707 \\ l_{2}&=111.802 \end{aligned}

$\therefore \quad$ Easting of $S_{3}=500+70.707 \sin 45^{\circ}$ $=550 \mathrm{m}$

Q28GATE 2015MCQ

In a closed loop traverse of 1 km total length, the closing errors in departure and latitude are 0.3 m and 0.4m, respectively. The relative precision of this traverse will be:

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📖 Explanation

Relative precision $=\frac{1}{p / e}$ where p= Perimeter of traverse e= Closing error Closing error, $e=\sqrt{0.3^{2}+0.4^{2}}=0.5 \mathrm{m}$ Relative precision $=\frac{1}{\frac{1000}{0.5}}=\frac{1}{2000}$

Q29GATE 2014MCQ

Group I lists tool/instrument while Group II lists the method of surveying. Match the tool/instrument with the corresponding method of surveying.

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Q30GATE 2013MCQ

A theodolite is set up at station A and a 3 m long staff is held vertically at station B. The depression angle reading at 2.5 m marking on the staffis 6 $^{\circ}$ 10'. The horizontal distance between A and B is 2200 m. Height of instrument at station A is 1.1 m and R.L. of A is 880.88 m. Apply the curvature and refraction correction, and determine the R.L. of B (in m). __________

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📖 Explanation

RL of $\mathrm{A}=880.88 \mathrm{m}$ RL of plane of collimation $=880.88+1.1=881.98 \mathrm{m}$

\begin{aligned} \frac{B B^{\prime \prime}}{C B^{\prime}} &=\tan 6^{\circ} 10^{\prime} \\ B^{\prime} B^{\prime} &=2200 \mathrm{m} \times \tan 6^{\circ} 10^{\prime} \\ &=237.701 \mathrm{m}\\ \text{RL of }B&=R L\text{ of Ine of sight }-B B^{\prime \prime} \\ &=881.98-237.701-(25-0.3257) \\ &=642.105 \mathrm{m} \end{aligned}

Q31GATE 2013MCQ

The latitude and departure of a line AB are +78 m and -45.1 m, respectively. The whole circle bearing of the line AB is:

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📖 Explanation

since the latitude of line is positive and departure is negative, the line lies in the second quadrant.

\begin{aligned} \therefore \quad l \cos \theta &=78 \\ l \sin \theta &=-45.1 \\ \Rightarrow \quad \tan \theta &=-0.578 \\ \quad \theta&=-30^{\circ}\\ \therefore W C B \text{ of } A B&=360^{\circ}-30^{\circ} \\ &=380^{\circ} \end{aligned}

Q32GATE 2013MCQ

Following bearings are observed while traversing with a compass. After applying the correction due to local attraction, the corrected fore bearing of line BC will be:

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📖 Explanation

The difference between back bearing and fore bearing of line DE $=258^{\circ} 30^{\circ}-78^{\circ} 30^{\circ}=180^{\circ}$ $\therefore$ Station D and E are free from local attraction Fore bearing of line $E A=216^{\circ} 30^{\circ}$ $\therefore$ Correct back bearing of line EA

\begin{array}{c} =216^{\circ} 30^{\circ}-180^{\circ}=36^{\circ} 30^{\circ}\\ \therefore \text{ Error at } A=31^{\circ} 45^{\circ}-36^{\circ} 30^{\circ} \\ =-4^{\circ} 45^{\circ} \end{array}

Hence correction at $A=+4^{\circ} 45$ $\therefore$ Correct fore bearing of line AB

\begin{array}{l} =126^{\circ} 45^{\prime}+4^{\circ} 45^{\prime} \\ =131^{\circ} 30^{\prime} \end{array}

Correct back bearing of AB

\begin{array}{l} =131^{\circ} 30+180^{\circ} \\ =311^{\circ} 30^{\prime} \end{array}

$\therefore$ Error at $B=-3^{\circ} 30^{\circ}$ Correction at $\mathrm{B}=+3^{\circ} 30^{\circ}$ Hence correct bearing of BC

\begin{array}{l} =45^{\circ} 15^{\prime}+3^{\circ} 30^{\prime} \\ =48^{\circ} 45^{\prime} \end{array}

Q33GATE 2011MCQ

The observations from a closed loop traverse around an obstacle are What is the value of the missing measurement (rounded off to the nearest 10 mm)?

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📖 Explanation

In a closed loop travers, the algebraic sum of all the latitudes should be equal to zeroi.e., $\Sigma.=0$

\begin{aligned} \Rightarrow\quad L \cos 33.7500^{\circ}&+300 \cos 86.3847^{\circ} \\ +354.524 \cos &169.3819^{\circ}+450 \\ \cos 243.900^{\circ}&+268 \cos 317.5^{\circ}=0 \\ \Rightarrow\quad L \cos 33.75^{\circ} &=329.9166 \\ \Rightarrow\quad L &=396.79 \mathrm{m} \end{aligned}

Q34GATE 2010MCQ

The local mean time at a place located in longitude 90 $^{\circ}$ 40' E when the standard time is 6 hours and 30 minutes and the standard meridian is 82 $^{\circ}$ 30' E is

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📖 Explanation

Difference between standard meridian and the longitude of the given $\text { place }=90^{\circ} 40^{\circ} \mathrm{E}-82^{\circ} 30^{\circ} \mathrm{E}=8^{\circ} 10^{\prime}$ The longitude of the given place is more than the standard meridian. Hence the local mean time of the given place will be ahead of the standard time. $8^{\circ}=\frac{8}{15} \text { hours }$ =0 hours and 32 minutes and 0 seconds $10^{\prime}=\frac{10}{15} \text { minutes }$ =0 hours and 0 minutes and 40 seconds Time for a difference of $8^{\circ} 10^{\prime}=32$ minutes and 40 seconds Local mean time = standard time +32 minutes and 40 seconds

\begin{aligned} =& 6 \text { hours and } 30 \text { minutes } \\ &+32 \text { minutes and } 40 \text { seconds } \\ =& 7 \text { hours, } 02 \text { minutes and } 40 \text{seconds} \end{aligned}

Q35GATE 2009MCQ

The magnetic bearing of a line AB was $N\; 59^{\circ} {30}' \; W$ in the year 1967, when the declination was $4^{\circ}10' E$ . If the present declination is $3^{\circ}\; W$ , the whole circle bearing of the line is

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📖 Explanation

True Bearing of

\begin{aligned} \mathrm{AB}&=\text { Magnetic bearing of } \mathrm{AB}\\ &- \text{ Magnetic declination}\\ &=\mathrm{N} 59^{\circ} 30 \mathrm{W}-4^{\circ} 10^{\prime} \\ &=\mathrm{N} 55^{\circ} 20 \mathrm{W} \end{aligned}

Present Magnetic bearing of

\begin{aligned} \mathrm{AB}&=\text { True bearing } \mathrm{AB}\\ &-\text{ Magnetic declination}\\ &=\mathrm{N} 55^{\circ} 20 \mathrm{W}-3^{\circ} \\ &=\mathrm{N} 52^{\circ} 20 \mathrm{W} \end{aligned}

Whole circle bearing of $A B=360^{\circ}-52^{\circ} 20^{\prime}=307^{\circ} 40^{\prime}$

Q36GATE 2009MCQ

In quadrantal bearing system, bearing of a line varies from

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Q37GATE 2008MCQ

The lengths and bearing of a closed traverse PQRSP are given below. The missing length and bearing, respectively of the line SP are

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📖 Explanation

For a closed traverse sum of latitutes and departures should be zero respectively l.e.

\begin{array}{l} \sum L=0\\ 200 \cos 0^{\circ}+1000 \cos 45^{\circ}+907 \cos 180^{\circ} \\ +L\cos \theta=0 \quad\ldots(i)\\ L\cos \theta=-0.10678 \\ 2D=0 \\ 200 \sin \theta^{2}+1000 \sin 45^{\circ}+907 \sin 180^{\circ} \\ +L\sin \theta=0 \\ L \sin \theta = -707.10678 \quad\ldots(ii) \end{array}

Dividing (i) by (ii) we get $\tan \theta = 6622.09$ $\theta = 270^{\circ}$ $\therefore L = \frac{-707.10678}{\sin 270^{\circ}}=707.10678m$

Q38GATE 2007MCQ

The following table gives data of consecutive coordinates in respect of a closed theodolite traverse PQRSP. The magnitude and direction of error of closure in whole circle bearing are.

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📖 Explanation

$\Delta L=400.75+100.25-199.0-300.0=2$ $\Delta D=-300.5+199.25+299.75^{*}-200.5=-2$ [* This value is given as 399.75 in the original question paper Which is incorrect] $\therefore$ Error of closure.

\begin{aligned} e &=\sqrt{(\Delta L)^{2}+(\Delta D)^{2}} \\ &=\sqrt{(2)^{2}+(-2)^{2}}=2.82 \mathrm{m}\\ \therefore \quad \tan \theta&=\frac{\Delta D}{\Delta L} \\ \Rightarrow \quad \theta&=\tan ^{-1}\left(\frac{-2}{2}\right) \\ \Rightarrow \quad \theta&=-45^{\circ} \\ \text{or}\quad \theta&=360-45=315^{\circ} \end{aligned}

Q39GATE 2007MCQ

The magnetic bearing of a line AB is S 45 $^{\circ}$ E and the declination is 5 $^{\circ}$ West. The true bearing of the line AB is

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📖 Explanation

The bearng of AB, $(\alpha)$ = MB of AB + Magnetic declination $=45^{\circ}+5^{\circ}=S\; 50^{\circ} \mathrm{E}$

Q40GATE 2006MCQ

The observed magnetic bearing of a line OE was found to be 185 $^{\circ}$ . It was later discovered that station O had a local attraction of +1.5 $^{\circ}$ . The true bearing of the line OE, considering a magnetic declination of 3.5 $^{\circ}$ E shall be

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📖 Explanation

Observed magnetic bearing of $\mathrm{OE}=185^{\circ}$ Local attraction at $0=+1.5^{\circ}$ Corrected magnetic bearing of $O E=185^{\circ}-1.5^{\circ}=183.6^{\circ}$ $\therefore$ True bearing of OE = Corrected magnetic declination of OE + Magnetic declination $=103.6^{\circ}+3.5^{\circ}=187^{\circ}$