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**Question 1: The data from a closed traverse survey PQRS (run in the clockwise direction) are given in the tableThe closing error for the traverse PQRS (in degrees) is ________. [2019 : 1 Mark, Set-II]**

= (2n - 4) x 90

= (2 x 4 - 4) x 90

= 4 x 90 = 360°

Given sum of interior angles, = 88 + 92 + 94 + 89 = 363°

Then error in interior angle = 363 - 360 = 3°

The interior angle should not be less than 30°. In this way, triangle ‘Q’ & S having less angles (acute angle).

⇒ Q & S are ill-conditioned.

**The length and direction (whole circle bearing) of closure, respectively are [2018 : 2 Marks, Set-I] ****(a) 1 m and 90° ****(b) 2 m and 90° ****(c) 1 m and 270° ****(d) 2 m and 270°****Answer: (a)****Solution:**

∑L = W northing- E southing

= (101 + 419) - (437 + 83)

= 0

∑D = E Easting - E Westing

= (173 + 558) - (96 + 634)

= 1 m

∴ Length of closure

And, direction of closure,

Hence, θ lies in I quadrant and θ = 90°**Question 4: The observed bearings of a traverse are given below****The station(s) most likely to be affected by the local attraction is/are [2017 : 2 Marks, Set-I]****(a) Only R ****(b) Only S ****(c) Rand S ****(d) Pand Q****Answer: (a)****Solution: **

Station R is most likely to be affected.

Since difference of FB and BB for PQ and ST is 180°.

Hence, P, Q, S, T are free from local attraction.**Question 5: The reduced bearing of a 10 m long line is N30°E. The departure of the line is [2016 : 1 Mark, Set-II]****(a) 10.00 m****(b) 8.66 m****(c) 7.52 m****(d) 5.00 m****Answer: **(d)**Solution: **

The departure of the line,

D = l sin θ

= 10 sin 30°

= 10/2 = 5m**Question 6: The bearings of two inaccessible stations, S _{1} (Easting 500 m, Northing 500 m) and S_{2} (Easting 600 m, Northing 450 m) from a station S_{3} were observed as 225° and 153° 26' respectively. The independent Easting (in m) of station S_{3} is: [2015 : 2 Marks, Set-II]**

Let S_{1}S_{3} = l_{1} and S_{2}S_{3} = l_{2} **Northing of S _{3}:**

500 + l

l

500 + l

l

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

l

l

∴ Easting of S

= 550m

Magnetic FB of AB = N 79°50' E

To find local attraction at station A

As station O is free from local attraction

Hence FB of OA will be correct,

Correct FB of OA = N 50°20' W ≌ 309°40'

∴ Correct BB of OA = 129°40'

∵ Obserbed FB of AO

= Observed BB of OA

= S52°40'E

= 127°20

Error = Observed bearing - Correct bearing

= 127°20' - 129°40'

= 2°20'

Correction = +2°20'

Local attraction at station A

= +2°20' ≌ 2°20' E

∴ Magnetic FB of AB = N 79°50'E

δ = 2°E and local attraction = 2°20'E

∴ TB of FB of AB = 79°50' + 2°20' + 2°

= N84°10'E

where, p = Perimeter of traverse

e = Closing error

D = kS (k = stadia constant)

RL of plane of collimation

= 880.88 + 1.1 = 881.98 m

Here, D = Distance between theodolite and staff in km

∴ RL of B = 881.98-237.701

- (2.5 x 0.0673 x 2.2

= 642.105 m

= 258° 30' - 78° 30' = 180°

∴ Station D and E are free from local attraction

Fore bearing of line EA = 216° 30'

∴ Correct back bearing of line EA

= 216° 30' - 180° = 36° 30'

∴ Error at A = 31° 45' - 36° 30' = -4°45'

Since the latitude of line is positive and departure is negative, the line lies in the fourth quadrant.

∴ L*cosθ = 78

L*sinθ = -45.1

⇒ tanθ = -0.578

θ = -30°

∴ WCB of AB = 360° - 30°

= 330°

**What is the value of the missing measurement (rounded off to the nearest 10 mm)? [2011 : 2 Marks](a) 396.86 m (b) 396.79 m (c) 396.05 m (d) 396.94 m**

⇒ Lcos 33.7500°+ 300 cos 86.3847° + 354.524 cos 169.3819° + 450 cos 243.900° + 268 cos 317.5° = 0

⇒ Lcos 33.75° = 329.9166

⇒ L = 396.79 m

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