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**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 Simpsonâ€™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 observerâ€™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 Î”BET,

tan 16.5Â° = x/y

â‡’ y = 3.3759x .....(i)

In Î”AOT,

......(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,

â‡’ **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,

âˆ‘F.S. > âˆ‘B.S.

âˆ´ Fall = âˆ‘F.S. - âˆ‘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 3Â° and 5Â° above horizontal, respectively. The vertical angles from P and Q to the base of the tower are 0.1 Â° and 0.5Â° 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) (tan5Â° + tan0.5Â°)

â‡’ x = 228.758

Height of tower= 228.758 (tan 3Â° + tan0.1Â°)

= 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 earthâ€™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 earthâ€™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.

âˆ´ 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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