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# Levelling and Contouring Civil Engineering (CE) Notes | EduRev

## Topic wise GATE Past Year Papers for Civil Engineering

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## Civil Engineering (CE) : Levelling and Contouring Civil Engineering (CE) Notes | EduRev

The document Levelling and Contouring Civil Engineering (CE) Notes | EduRev is a part of the Civil Engineering (CE) Course Topic wise GATE Past Year Papers for Civil Engineering.
All you need of Civil Engineering (CE) at this link: Civil Engineering (CE)

Question 1: A series of perpendicular offsets taken from a curved boundary wall to a straight survey line at an interval of 6 m are 1.22,1.67, 2.04, 2.34, 2.14, 1.87, and 1.15 m. The area (in m2, round off to 2 decimal places) bounded by the survey line, curved boundary wall, the first and the last offsets, determined using Simpson's rule, is__________.    [2019 : 2 Marks, Set-II]
Solution: Area by Simpson’s rule Question 2: A staff is placed on a benchmark (BM) of reduced level (RL) 100.000 m and a theodolite is placed at a horizontal distance of 50 m from the BM to measure the vertical angles. The measured vertical angles from the horizontal at the staff readings of 0.400 m and 2.400 m are found to be the same. Taking the height of the instrument as 1.400 m, the RL (in m) of the theodolite station is_______.    [2019 : 2 Marks, Set-I]
Solution: HI = RL of BM + BS
= 100 + (0.4 + 1)
= 101.4
RL of Theodolite station
= Hl-height
= 101.4 - 1.4
= 100 m

Question 3: A level instrument at a height of 1.320 m has been placed at a station having a Reduced Level (RL) of 112.565 m. The instrument reads -2.835 m on a levelling staff held at the bottom of a bridge deck. The RL (in m) of the bottom of the bridge deck is    [2018 : 2 Marks, Set-II]
(a) 116.720
(b) 116.080
(c) 114.080
(d) 111.050
Solution: RL of bottom of bridge deck
= 112.565 + 1.320 + (2.835)
= 116.720 m

Question 4: An observer standing on the deck of a ship just sees the top of a lighthouse. The top of the lighthouse is 40 m above the sea level and the height of the observer’s eye is 5 m above the sea level. The distance (in km, up to one decimal place) of the observer from the lighthouse is _____. [2017 : 2 Marks, Set-II]
Solution: Distance of observer from the lighthouse, Question 5: The vertical angles subtended by the top of a tower T at two instrument stations set up at P and Q, are shown in the figure. The two stations are in line with the tower and spaced at a distance of 60 m. Readings taken from these two stations on a leveling staff placed at the benchmark (BM = 450.000 m) are also shown in the figure. The reduced level of the top of the tower T (expressed in m) is__________    [2016 : 2 Marks, Set-I Solution: In ΔBET,
tan 16.5° = x/y
⇒ y = 3.3759x  .....(i)
In ΔAOT, ......(ii)
Put value of y in eg. (ii) 11.1203 + 0.6255x = x + 2
x = 24.3568 m
So, the reduced level of tower,
T = 450.000 + 2.555 + 24.356

= 476.911 m

Question 6: The staff reading taken on a workshop floor using a level is 0.645 m. The inverted staff reading taken to the bottom of a beam is 2.960 m. The reduced level of the floor is 40.500 m. The reduced level (expressed in m) of the bottom of the beam is    [2016 : 1 Mark, Set-I]
(a) 44.105
(b) 43.460
(c) 42.815
(d) 41.145
Solution: Height of instrument, HI
= R.L. of floor + staff reading from floor
= 40.500 + 0.645
= 41.145 m
R.L. of bottom of beam
= HI + inverted staff reading taken from bottom of beam
= 41.145 + 2.960
= 44.105 m

Question 7: Two pegs A and B were fixed on opposite banks of a 50 m wide river. The level was set up at A and the staff readings on Pegs A and B were observed as 1.350 m and 1.550 m, respectively. Thereafter the instrument was shifted and set up at B. The staff readings on Pegs B and A were observed as 0.750 m and 0.550 m, respectively. If the R.L. of Peg A is 100.200 m, the R.L. (in m) of Peg B is _____.    [2015 : 2 Marks, Set-II]
Solution: Staff readings shows that station B is below station A.
a1 = 1.350
a2 = 0.550
b3 = 1.550
b2 = 0.750 = 0.200
RL of B = RL of A - 0.200
= 100.200-0.200
= 100.000 m

Question 8: The combined correction due to curvature and refraction (in m) for distance of 1 km on the surface of Earth is    [2015 : 1 Mark, Set-II]
(a) 0.0673
(b) 0.673
(c) 7.63
(d) 0.763
Solution: Correction due to curvature,
Cc = -0.0785 d2
Correction due to refraction,
Cr = +0.0112 d2
Composite correction,
C = -0.0785d2 + 0.0112 d2
= -0.0673 d2
where d is in km, C is in m, Question 9: In a leveling work, sum of the Back Sight (B.S.) and Fore Sight (F.S.) have boon found to be 3.085 m and 5.645 m respectively. If the Reduced Level (R.l .) of the starting station is 100.000 m, the R.L. (in m) of the last station is    [2015 : 1 Mark, Set-II]
Solution: Using Rise and Fall method,
∑F.S. > ∑B.S.
∴ Fall = ∑F.S. - ∑B.S. = R.L. of first station - R.L. of last station = 5.645-3.085 = 2.56 m
R.L. (last station) = R.L. (first station) - Fall
= 100 - 2.56 = 97.44 m

Question 10: Which of the following statements is FALSE?    [2015 : 1 Mark, Set-I]
(a) Plumb line is along the direction of gravity
(b) Mean Sea Level (MSL) is used as a reference surface for establishing the horizontal control
(c) Mean Sea Level (MSL) is a simplification of the Geoid
(d) Geoid is an equi-potential surface of gravity
Solution: Mean sea level (MSL) is used as a reference surface for establishing the vertical control.

Question 11: A levelling is carried out to establish the Reduced Levels (RL) of point R with respect to the Bench Mark (BM) at P. The staff readings taken are given below:    [2014 : 2 Marks, Set-I] If RL of P is +100.000 m, then RL (in m) of R is
(a) 103.355
(b) 103.155
(c) 101.455
(d) 100.355
Solution:
HI = RL + BS
and RL = HI - FS Question 12: The Reduced Levels (RLs) of the points Pand Q are +49.600 m and +51.870 m respectively. Distance PQ is 20 m. The distance (in m from P) at which the +51.000 m contour cuts the line PQ is    [2014 : 1 Mark, Set-I]
(a) 15.00
(b) 12.33
(c) 3.52
(d) 2.27
Solution:  Question 13: The horizontal distance between two stations P and Q is 100 m. The vertical angles from P and Q to the top of a vertical tower at T are 3° and 5° above horizontal, respectively. The vertical angles from P and Q to the base of the tower are 0.1 ° and 0.5° below horizontal, respectively. Stations P, Q and the tower are in the same vertical plane with P and Q being on the same side of T. Neglecting earth's curvature and atmospheric refraction, the height (in m) of the tower is    [2012 : 2 Marks]
(a) 6.972
(b) 12.387
(c) 12.540
(d) 128.745
Solution: x (tan 3 + tan 0.1) = (x - 100) (tan5° + tan0.5°)
⇒ x = 228.758 Height of tower= 228.758 (tan 3° + tan0.1°)
= 12.387 m
Note: Question has erroreous data, hence many other answers are also possible.

Question 14: Which of the following errors can be eliminated by reciprocal measurements in differential leveling?    [2012 : 1 Mark]
I. Error due to earth’s curvature
II. Error due to atmospheric refraction
(a) Both I and II
(b) I only
(c) II only
(d) Neither I nor II

Solution:
Reciprocal levelling eliminates the errors due to
(ii) Combined effect of earth’s curvature and the refraction of the atmosphere.
(iii) Variation in the average refraction.

Question 15: Curvature correction to a staff reading in a differential leveling survey is    [2011 : 1 Mark]
(a) always subtractive
(b) always zero
(d) dependent on latitude

Question 16: A bench mark has been established at the soffit of an ornamental arch at the known elevation of 100.0 m above mean sea level. The back sight used to establish height of instrument is an inverted staff reading of 2.105 m. A forward sight reading with normally held staff of 1.105 m is taken on a recently constructed plinth. The elevation of the plinth is    [2010 : 2 Marks]
(a) 103.210 m
(b) 101.000 m
(c) 99.000 m
(d) 96.790 m
Solution: Height of instrument = BM + Back sight
Since the staff is inverted, the back sight will be negative.
∴ Height of instrument
= 100 + (- 2.105) = 97.897 m
Elevation of plinth = Height of instrument- Fore sight
= 97.895-1.105 = 96.790 m

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