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)

**Q****uestion 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 m ^{2}, 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]**

Area by Simpson’s rule

HI = RL of BM + BS

= 100 + (0.4 + 1)

= 101.4

RL of Theodolite station

= Hl-height

= 101.4 - 1.4

= 100 m**Q****uestion 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****Answer: (a)****Solution:**

RL of bottom of bridge deck

= 112.565 + 1.320 + (2.835)

= 116.720 m**Q****uestion 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,**Q****uestion 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**Q****uestion 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 ****Answer: (a)****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**Q****uestion 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.

a_{1} = 1.350

a_{2} = 0.550

b_{3} = 1.550

b_{2} = 0.750

= 0.200

RL of B = RL of A - 0.200

= 100.200-0.200

= 100.000 m**Q****uestion 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****Answer: **(a)**Solution: **Correction due to curvature,

C_{c} = -0.0785 d^{2}

Correction due to refraction,

C_{r} = +0.0112 d^{2}

Composite correction,

C = -0.0785d^{2} + 0.0112 d^{2}

= -0.0673 d^{2}

where d is in km, C is in m,

⇒ **Q****uestion 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**Q****uestion 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****Answer: (b)****Solution: **Mean sea level (MSL) is used as a reference surface for establishing the vertical control.**Q****uestion 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****Answer: (c)****Solution: **

HI = RL + BS

and RL = HI - FS**Q****uestion 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****Answer: **(b)**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****Answer: (b)****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 curvatureII. Error due to atmospheric refraction(a) Both I and II(b) I only(c) II only(d) Neither I nor II**

Reciprocal levelling eliminates the errors due to

(i) Error in instrument adjustment.

(ii) Combined effect of earth’s curvature and the refraction of the atmosphere.

(iii) Variation in the average refraction.

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

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!