TEST PAPER OF SURVEYING.docx

Sub: Surveying Topic : Basics of surveying

1. The area of a plan of an old map plotted to a scale of 10 m to 1 cm measures 100.2

cm2 as measured by a planimeter. The

plan is found to have shrunk so that line originally 10 cm long now measures 9.7 cm. further; the 20 m chain used was 8 cm too short. Find the true area of survey. [Ans: 105.6434 acres]

2. An area actually measures 0.8094

hectares. How much will it measure in m2

by a 30.48 m chain which was 20.32 cm too short at the start and 60.96 cm too long at the end of the survey?

[Ans: 7987.15 m2]

3. The area of apiece of a land which had been surveyed with a chain was calculated

to be 9562 m2. Of this, 8935 m2 was the

total area of the triangles and 627 m2 was

the area included between chain lines and the boundary. The 30 m chain used was found 0.05 m too long, and the 30 m tape used for measuring offsets was found 0.03 m too short from their nominal lengths. Calculate the correct area of the land.

[Ans: 9590.5 m2]

4. A line measured with a steel tape which

was exactly 30 m at a temperature of 240

C and a pull of 10 kg. the measured length was 1650 m. the temperature during

measurement was 300C and a pull applied

was 15 kg. find the true length of the line, if the cross sectional area of the tape was

0.025 cm2. The coefficient of thermal

expansion of the material of the tape per0

C is 3.5×10−6 and the modulus of elasticity of the material of tape is 2.1×106 kg/cm2.

[Ans: 27.7768 m]

5. The slope distance between two stations A and B of elevations 1572.25 m and 4260.46 m, corrected for meteorological conditions is 33449.215 m. determine sea level distance, R = 6370 km. [Ans: 33332.789 m]

6. The following are the observed fore bearings of the line. Calculate the respective back bearings.

(i) AB 24030 '

(ii) CD 1200

(iii) EF 2680 30'

(iv) GH 3530

(v) AB −N 620E

(vi) CD S 12030 'E

(vii) EF S 590 30'W

(viii) GH N 880 E

[Ans: (i) AB 204030 '

(ii) CD 3000

(iii) EF 88030'

(iv) GH 1730

(v) AB S 620W

(vi) CD N 12030 ' W

(vii) EF N 590 30'E

(viii)GH S 880 E

7. Convert the following whole circle bearings to quadrantal bearings.

(i) 100 18'

(ii) 95012'

(iii) 2250 30'

(iv) 3500 10'

[Ans:N 100 18' E,S 840 48' E,

S 45030' W ,N 90 50' W ]

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8. Convert the following quadrantal bearings to whole circle bearings.

(i) N 60 20'E

(ii) S 170 25' E(iii) N 170 25' W(iv) N 110 W

[Ans:

6020' ,162035' ,342035' , 3490 00']

9. Find the value of magnetic declination if the magnetic bearing of the sun at noon is

2560?

[Ans: 40E]

10. A line was drawn to a magnetic bearing of

S50000'E on old plane when the magnetic

declination as 2040'W. to what bearing

should it be set now if the present

declination is 3020'E?

[Ans: S560E]

11. The bearing of the sides of a triangle ABC are as follows. Computer the interior angles.

AB = 600, BC = 1300, and CA= 2700

[Ans:

12. Find out the bearing of the lines of an equilateral triangle ABC running clockwise, if the bearing of the line AB is

600 30' .

[Ans: BC = 180033' , CA = 3000

30'

13. The following bearing were observed in running a compass traverse:Line F.B. B.B.

AB 66015' 244000'

BC 129045' 313030'

CD 218030' 37030'

DA 306045' 126045'

Find the correct fore and back bearings and the true bearings o the lines, given

that the magnetic declination is 1040'E.

[Ans: True bearing AB = 67055',

BC = 133040' , CD = 219010' ,

DA = 308025']

14. The fore and back bearings of a closed traverse conducted at Naini, Allahabad are given below. Indicate which stations are affected by local attraction. Also find out the corrected bearings. If the value of

declination is 40W, find out the true

bearings.Line Fore bearing Back bearing

AB S550 30'E N550 30'W

BC N680 15'E S66000'W

CD N49030 'W S440 45'W

DA S200 15'W N170 45 'E

[Ans:AB=S590 30'E,BC=N

64015 'E, CD= N510 15'W,

DA= S130 45 ' W ]

15. To find out the included angles in a closed survey PQRSTP, following observations were made with the compass. Calculate the included angles after correction for local attractions.Line Fore bearing back bearing

PQ N620 45' E S620 15' W

QR N210 00' E S200 45 ' W

RS N 71030 ' W

N 71030 ' E

ST S390 00' W

N 380 00' E

TP S 540 30' W

N 530 15' W

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[Ans: TPQ=

115054 ' , PQR=1380 390 ,QRS=870 39' ,RST=110024 ' , STP=87024 '

]

16. If the fore bearing of a line is zero degree, its back bearing is

a) 00 b) 1800 c) 3600 d) 2700

17. Which will be included angle AOB if the bearings of the lines AO and OB are respectively 400 and 1300?

a) 900 b) 1700 c) 2700 d) 1300

18. Which will be the included angle AOB if the bearings of the lines AO and OB are respectively 400 and 1300?

a) 900 b) 1700 c) 2700 d) 1300

19. If the magnetic bearing of a line is 34030'

and the magnetic declination 30 30' W ,

the true bearing of the line will be

a) 370 b) 380 c) 300 d) 310

20. The force bearing of line AB is 2090. The

included angle ABC is 3410. The F.B. of

the line BC is

a) 5500 b) 3300 c) 100 d) 1900

21. The magnetic bearing of the sun at noon is

1780. The magnetic declination at the

place is a) 20 W

b) 20 E

c) 20 N

d) 20S

22. The bearings to two inaccessible stations A and B taken from a station C were

2500 00' and 1530 26' , respectively. The

coordinates of A and B were as follows: Station Easting Northing A 300 200 B 400 150Calculate the independent coordinates of C. [Ans: 363.45, 223.10]

23. A closed transverse ABCD in which the bearing of AD has not been observed and

the length of BC has been missed out in recording, was conducted at Allahabad. The rest of the field record is as follows: Line Bearing Length (m)

AB 1810 18' 335

BC 90000' -

CD 357036' 408

DA - 828 Calculate the missing bearing and the length.

[Ans: Bearing =264057 ' 40 ¿, Length =

849.5 m]

24. A straight line AC of length 2000 m is required to be sent out at right angles to a given line AB. This is done by traversing from A towards C. the observations recorded are as follows: Line Length (m) Bearing

AB - 3600 00'

AD 731 1130 48'

DE 467 810 18'

EF 583 1050 54'

Calculate the necessary length and bearing of PC. [Ans: length= 492.85 m, Bearing =

380 48 ' 13 ¿]

25. A helicopter files in sky from Manipuri(M) in Allahabad to Teliarganj (T) as per following conditions:

a) 5.0 km along a 50 00' up gradient

eastwards up to B

b) 3.0 km along an30 00' up gradient

northwards up to C

c) 4.0 km along an 40 00' up gradient in

the N-W direction up to D

d) 4.0 km along a 40 00' down gradient

south wards up to M calculate the bearing, distance and gradient to reach the point T. the starting point M has coordinates N-100 m and E-200 m and its height above datum is 100 m. [Ans: Bearing of TM=

S 490 45' 52} W, horizontal distance MT=2828.7 m, gradient=1 in 4.77 downward ¿

]

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26. In a transverse ABCDEFG, the line BA is taken as the reference meridian, the coordinates of the sides AB, BC, CD,DE,EF are:

Line AB BC CD DE EFNorthing

-1190.9

-565.3

590.5

606.9

1017.2

easting

0 736.4

796.8

-468.0

370.4

If the bearing of FG is 284013 ' and its

length is 896.00m, find the length and bearing of GA.

[Ans: 884.21 m, 390 53' 19 ¿

27. For a railway project, a tunnel is to be run between two points P and Q points coordinates N EP 0 0 Q 4020 800 R 2120 1900It is desired to sink a shaft at S, the mid-point of PQ. S is to be fixed from R, the third known point. Calculate (i) the coordinates of S (ii) the length RS (iii) the bearing of RS [Ans: N-2010,E-400; 1504.028 m;

2650 48 ']

28. A surveyor is running a line in the

direction AB having a bearing N100 30'

W. It was impossible to continue the line because of an obstruction. The obstruction was being transverse random as ABCDE and it was continued. What should be the distance DE so that points A, B and E lie on a straight line? What deflection angle must be turned at E in order to continue the line AB? Also, calculate the distance EB. The data observed was, Point deflection line Length (m) Angles

B 620 08' (R) BC 120.00

C 780 16' (L) CD 280.00

D 800 40'( L)

[Ans: 28.48 m,83012' R ,321.69 m]

29. For a closed ABCDA, compute the missing data.

Line Length (m) bearingAB 100.00 N45030 ' WBC 605.00 N45030 ' WCD 95.00 N88020' EDA ? ?

[Ans: 679.97 m; 1860 53' 35. 5 ¿

]

30. In a quadrilateral ABCD, the coordinates of the points (in metres) are as follows:

point E NA 100 100B 208 104C 223 353D 100 357

Calculate the area.

[Ans: 29221.5m2]

31. The following observations were made for a closed transverse ABCDEA.

Line Length (m) Included angles

AB 1512.1BC 863.7CD ?DE ?EA 793.7

It was not possible to occupy D, but it could from C and E. calculate the observations that could not be made taking DE as datum assuming all the observations to be correct.

[Ans: 106045' ,DE = 1776.89 m,

CD = 1170.96 m]

32. A straight road AB is proposed to be extended in the direction AB produced,

the c/l of the road is obstructed by a farm.

A point is to be fixed beyond the form so that A, B and C lie in a straight line and a traverse is run from B to C. The following observations were made:

ABD = 8704 2' BD = 29.02 m

BDE = 2820 36' DE = 77.14 m

DEC = 2910 0 6'

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Calculate

a) the length of lime EC.b) the angle to be measured at C, so that

the c/l of the road can be extended beyond

C. (c) the chainage of C assuming the chainage of A to be 100 m and AB to be 130.64 m.

[Ans: 17.82 m, 580 36' ,296.07 m]

33. Find the error of reading on a level staff, if the observed reading is 3.830 m and at the point sighted on the staff is 0.15 m off the vertical through the bottom. [ Ans: 0.0029]

34. The following staff readings were taken with a level. The instrument having been

shifted after the 4 th , 7 th∧10 th readings.

R.L of the starting B.M. is 100.00 m. enter the readings in the form of a level book page and reduce the level by the collimation method and apply the usual checks. 2.65, 3.74, 3.83, 5.27, 4.64, 0.38, 0.96, 1.64, 2.84, 3.48, 4.68 and 5.26. [Ans: 100.00, 98.91, 98.82, 97.38, 101.64, 101.06, 99.86, 99.22, 98.64]

35. During fly leveling the following note is made. B.S:0.62, 2.05, 1.42, 2.63 and 2.42 m F.S: 2.44, 1.35, 0.53 and 2.41 m The first B.S. was taken on a B.M. of R.L. 100.00 m. from the last B.S. it is required to set 4 pegs each at a distance of 30 m on a rising gradient of 1 in 200. Enter these notes in the form of a level book and calculate the R.L. of the top of each peg by the rise and fall method. Also, calculate the staff readings on each peg by the rise and fall method. Also, calculate the staff readings on each peg. [Ans: R.L. 100.00, 98.18, 98.88, 99.77, 99.99, 100.14, 100.29, 100.44, 100.59, staff readings on pegs 2.27, 2.12, 1.97, 1.82 m]

36. The reduced level of ground at four points A, B,C and D are 54.35, 54.30, 54.20,

54.30 m, respectively. A sewer is to be laid so that its invert is 3.048 m below the ground at A and it falls with a uniform gradient of in 340 to D. the distances AB, AC and AD are 35.845, 80.742 and 134.7 m, respectively. Find the invert level and depth of trench at B, C and D. [Ans: level 51.1966, 51.065, 50.906 m. depth 3.103, 3.135, 3.394 m, (negative)

37. Reciprocal leveling was done between two points A and B situated on the opposite sides of a valley 730 m wide. The following data was collected:

Instrument at

Height of instrument

Staff at

Staff reading (m)

A 1.463 B 1.688 B 1.463 A 0.991

Determine the difference in level between A and B and the same amount of collimation error if any. [Ans: 0.3485 m (fall from A to B, 0.1594 m (negative)]

38. The following consective readings were taken with a dumpy level and 4 m leveling staff on a continuously sloping ground at 30 m intervals. 0.680, 1.455, 1.855, 2.885, 3.380, 1.055, 1.860, 2.265, 3.540, 0.835, 0.945, 1.530 and 2.250. The R.L. of starting point 80.750 ma) rule out a page of the level book and enter the above readings. (b) carry out reduction of heights by the collimation method and apply the usual checks. c) determine the gradient of the line joining the first and last points.[Ans: R.L.: 80.750, 79.975, 79.100, 78.050, 77.245, 76.840, 75.565, 75.455, 74.870, 74.150; gradient, 1 in 50 fall]

39. In leveling between the two points A and B on opposite banks of a river, the level was set up near A and the staff readings on A and B and 2.150 m and 3.565 m, respectively. The level was then moved to B and the respective staff readings on A and B were 1.965 m and 3.260 m. find the true difference in levels of A and B. [Ans: 1.355 m fall]

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40. Calculate the combined correction for curvature and refraction for distance of (i) 5 km and (ii) 500 m. [Ans:1.6835 m, 0.0168 m]

41. A level is set up at C on a line AB at 60 m from A and 700 m from B. The B.S. on A is 2.650 m and the F.S. on B is 2.780 m. find the difference in levels of A and B. [Ans: 0.097 m fall]

42. The following staff readings in meters were obtained when leveling along the centre-line of a straight road ABC using a digital level.

B.S I.S F.S Remarks2.405 Point

A(RL=250.05 m)

1.954 1.128 C.P0.169 1.466 Point B

2.408 Point D -1.515 Point E

1.460 2.941 C.P2.368 Point C

D is the highest point on the road surface beneath a bridge crossing over the road at this point and the staff was held inverted on the underside of the bridge grider at E, immediately above D. reduce the levels, correctly apply the checks, and determine the headroom at D. if the road is to be regarded so that AC is a uniform gradient, what will be the new headroom at D? the distance AD=240 m and DC= 60 m. [Ans: 3.923 m,5.071 m]

43. A line of levels was run in the form of a loop 650 m long. The initial starting point was 100.00 m. the last fore sight reading on the initial station was 2.465 m and the height of instrument was 102.0 m. are the levels acceptable for ordinary leveling? [Ans: No]

44. A line of levels was run from A to B. the leveling was then continued to a B.M. of elevation 40.0 m. the readings obtained are as follows. Obtain the reduced levels A and B

B.S I.S F.S R.L REMARKS

1.195 A0.445 2.3702.150 0.995

0.72 B1.465 0.2602.630 0.905

1.305 40.0 B.M(ANS: R.L of A =37.950M ,R.L. of B=97.655 M)

45. In a two-peg test of a dumpy level, the following readings were taken : Instrument at R Staff readings Midway between at P 1.525mpegs P and Q at Q 1.230mInstrument near P at P 1.420m At Q 1.025mHow much is the line of collimation inclined upwards or downwards? What should be the correct staff reading at Q with the instrument near P to keep the line of collimation truly horizontal ?[Ans: 0.10m, downwards, 1.125]

46. Two pegs A and B driven 100.0m apart. In the adjustment of a dumpy level, the following observations were recorded

INSTRUMENT NEAR

STAFF READINGS on A B R.L(m)

A 1.520 1.535 80.00B 1.595 1.58 ?

State whether the instrument is in adjustment. Find the R.L . of b.

(Ans: No, 80 m)

47. An instrument was set up at O and the angle of elevation to a vane 4 m above the

foot of the staff held at Q was 90030’.

The horizontal distance between the instrument and the staff was 2 km . determine the R.L. of the staff station O. the line of collimation was 2650 m .[Ans: 2984.95 m]

48. To find the elevation of the top F of a hill, a flag staff of 2 m height was erected and observation were made from two stations P and Q 60m apart . the horizontal angles

FPQ and FQP were 600 30 ' and 600 18 ' , respectively. Angle of elevation as observed at P and Q to the top F were

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100 12' and 100 48 ' respectively . the

staff readings on B.M. were 1.965m and 2.055m , respectively . when the instruments where at P and Q. the R.L of B.M is 430.00m . calculate the elevation of the top of the hill.(ans: 444.9m)

49. Find the reduced level of the top of a church apire from the following date

Instrument station

Reading on B.M

Vertical angle

Remark

A 1.578 100 12' R.L of B.M : 543.075m

B 1.269 8020 ' Distance AB =30m

Stations A and B the church spire are in the same vertical plane.(Ans: 566.628)

50. To determine the elevation of the top of a chimney the following observations were made:

Instrument station

Reading on B.M

Vertical angle

Remark

P 1.377m 110 53 ' R.L of B.M : 50.15m

Q 1.263m 805 'Stations P and Q the chimney are in the same vertical plane . Find the elevation of the chimney if the distance between P and Q was 30m(Ans: R.L.64.283m)

51. It was required to obtain the elevation of the top of a television tower located on the following data was obtained. A line AB, 135.0m long was staked out and the horizontal angles to the tower were

observed at A as 580 30 'and at B as 300.

At point B a back sight of 2.00m was taken on a B.M of elevation 100.00m and the vertical angle to the top of tower was

found to be 540. Calculate the elevation

of the tower (Ans: 256.284m)

52. A line of levels was run from a bench mark of R.L. 51.450 and ended on a B.M. of R.L. 63.500. the sum of the back sights and foresights were 87.775 and 73.725, respectively . what was the closing error of the work?

(Ans: 1.980)53. A dumpy level is set A and the staff

readings at A and a distant station B are 0.95m and 2.87m . when the instrument is set at B, the observed sights are 0.46m at A and 1.01m at B. if AB=300m, express the colearth and estimate the curvature error

(ans: 007' 44 ' ' )54. The top of a stack was sighted from two

stations A and B,which are 15m apart and are in the same vertical plane with the top of the stack. The observed from

instrument station A is 300 25 ' and that

observed from instrument station B is

220 28 ' . the angle of elevation from B to

a vane 1.75m above the foot of the staff

held at station A is 160 15 '. The heights

of the instrument at A and B were 1.856m and 1.565m respectively . the R.L of stations B is 100m. find out the R.L. of the top of the stack (Ans: 174.468m)

55. In order to determine the elevation of top Q of a signal on a hill ,observations were made from two stations P and R . the stations P,R, and signal Q were on the same plane . if the angles of elevation of the top Q of the signal measured at P and

R were 250 35 ' and 150 5 ' respectively,

determine the elevation of the foot of the signal if the height of the signal above it’s base was 43.The staff readings upon the bench mark (R.L. 105.42) were respectively 2.755 and 3.855m when the instrument was at P and at R . the distance between P and R was 120m.(Ans : 180.686m)

56. The following observations were made in running fly levels from a bench mark of R.L. 60.65:Back sight : 0.964,1.632,1.105,0.850Fore sight 0.948,1.153,1.984Five pegs at 20m intervals are to be set on a falling gradient of 1 in 100m from the last position of the instrument . the first peg is to be at R.L.60.Work out the staff readings required for setting the peg and prepare the page of the level book.

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(Ans: staff reading of peg 1:1.116m peg2:1.316m, peg3:1.516m,peg4:1.716m,peg5:1.916m)

57. Correction due to refraction is given bya) 0.0112D²b) 0.0785D²c) 0.0673D²d) 0.0012D²

58. What will be the correction for curvature for a distance of 1000m?a) 0.0673mb) 0.0785mc) 78.50md) 6.73m

59. The reading on B.M.=100m was 3.250M.the inverted staff reading to the bottom of a girder was 1.250m . the R.L of the bottom of girder isa) 101.250b) 102.0c) 104.50d) 103.250

60. The raeding on a 4.0m . staff at a pont is observed as 2.895m . if the staff was 8cm out of the plund line , the correct reading should have been a) 2.8938mb) 2.8150mc) 2.8961md) 2.8950 m

61. A level when set up 25m from peg A and 50m from peg B reads 2.847m on staff held on A and 3.462m on staff held on B , keeping the bubble at its center while reading. If the reduced levels of A and B are 283.665 m and 284.295m respectively. What is the collimation error per 100.0m?a) 0.015mb) 0.30mc) 0.045md) 0.060m

62. Following observations were recorded with a tachometer fitted with an analecticlens (K=100,C=0). Calculate the reduced level of change point at station T. the staff was held vertica; during the observations and the reduced level of B.M. was 500.m .

Instrument station

H.I Staff station

Vertical angle

Staffs readings

O 1.500 B.M -

40 30 '1.250,1.400,1.550

O 1.500 C.P6012 '

1.550,1.750,1.950

T 1.350 C.P −70 45'1.390,1.550,1.710

(Ans: 506.291 m, 510.767 m)63. The following readings were taken by a

tachometer from station B on stations A,C and D in clockwise directions

sight Horizontal circle reading

Vertical angle

Staff readings(m)

top middle bottom

A3010 10'

C1520 56 '−50 00 '

1.044 2.283 3.522

D2050 06 '+2. 30 '

0.645 2.376 4.100

The line BA has a bearing of 580 46 ' and

the instrument constants are 100 and 0. Find the slope of the line CD and its bearing.

(Ans: slope 1 in 7.54 rising, W.C.B.7026' 7 ' ')

64. A tacheometer has a multiplaying constant of 100 and an additive constant of 0. When set 1.35m above station B, the following readings were obtained

station sight Horizontal angle

Vertical angle

Staff readings(m)

A280 21 '

BC

820 03 ' 00 ' '200 30 '1.140,2.292,3.420

The coordinates of A are N00 and E163.86 while that of B are N118.41 and E163.86.Find the coordinates of C and its height above the datum, if the level of B is 27.30m (Ans: N90.74,E361.97,height=101.17 m)

65. A tacheometer is set up at an intermediate point on a traverse course PQ and the following observations were made on a vertically held staff:

Staff station Vertical angle Staff intercept point (m)

Axial hair readings (m)

P9030 '

2.250 2.105

Q6000 '

2.055 1.875

The instrument is fitted with an anallactic lens . the multiplying constant is 100. Computer the length PQ and R.L of Q if the R.L of P is350.50m(Ans: PQ=422.13m, R.L=335.46m)

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66. The following readings were taken by a tacheometer with the staff held vertical . the tacheometer is fitted with an anallatic lens and the multiplying constant is 100. Find out the horizontal distance from A to B and the R.L of B

Instrument station

Staff station

Vertical angle

Staff readings(m) Remarks

B.M −600 '1.100,1.153,2.060

R.L of B.M =976.00m

A B −800 '0.982,1.105,1.188

(ans: R.L=988.81m)67. A tacheometer is placed at station A and

readings on staff held vertical upon a B.M of R.L =100.00m and at a station B are 0.640,2.200, the telescope in the first case

is 6019 ' and in the second case is

070 42' . Find the horizontal distance

from A and B and the R.L of station B , if the instrument has constants 100 and 0.5(Ans: R.L=78.15m,D=414.92m)

68. Determine the gradient from a point P to a point Q and the distance PQ. Observations were made. With a tacheometer and the staff was held vertical at each of the stations . the instrument was made fitted with an anallactic lens

Instrument station

Staff station

bearing Vertical angle

Staffs reading(m)

P1340 100 32'

1.365,1.920,2.475

O Q2240 50 6 '

1.065,1.885,2.705

69. It was required to find the distance between two points A and B and their reduced levels. Two arbitrary points C and D were suitadly selected and the tacheometric observations recorded were as follows . the reduced levels of C and D were 100m and 110m, respectively

instrument

H.I(m)

Total coordinates

Staff station

Q.B Vertical angle

Staff reading(m)

C 1.450

N EA

N

390 24 ' WE160 24 '1.65,2.75,3.85

300.00

812.170

D 1.500

586.6500

1250.750

B

N 370 13 ' E25012 '2.50,3.20,3.90

Calculate :i) Length of line ABii) Gradient of line A/biii) Reduced levels of A and B

(Ans:673.83m,1in170.16(rising),1158.29m,162.24m)

70. The constant for an instrument are 1000 and 0.5. calculate the distance from the instrument to the staff when the micrometer readings are 5.246 and 50246. The staff intercept is 2.0m and the vertical angle measured is

+40 30 ' , the staff being held vertical

(Ans: 189.95 m)71. In the tangential method of tacheometry

two vanes were fixed 2m apart, the lower vane being 0.5m above the foot of the staff held vertical at station A. the vertical

angles measured were +10 12' and

−10 30. Find the horizontal distance and

R.L of A, if the height of the line of collimation is 100m(Ans: 42.3m ,98.388m)

72. Two targets spaced 6.0 m apart weer fixed on a subtense bar and the vertical angles measured on the two upper and lower

targets were 20 29 ' 30' 'and 10 28 ' 40 ' ', respectively. If the lower targets was at an elevation of 249.2m, what was the height of instrument(Ans: 240.43m)

73. The multiplaying constant of a tacheometer isa) f/ib) (f/d)+ic) (f/i)+dd) F+d

74. For a tacheometer equipped with an anallatic lens , the additive and multiplying constants are, respectivelya) 0 100b) 100 0c) 0 0d) 100 100

75. If the focal length of an object glass is 25cm, stadia interval is 1.25mm and the distance from object glass to the trunnion axis is 15cm, the additive constant isa) 0.1b) 0.4c) 1.66d) 20

76. If the intercept on vertical staff is observed as 0.75m from a tacheometer, with the line of sight horizontal, fitted with anallatic lens , the horizontal distance the tacheometer and the staff station is

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a) 0.75mb) 7.5mc) 75m d) 750m

77. In setting up the plane table at a station P, the corresponding plotted point p was not accurately centered over P. if the displacement of P was 10cm in a direction at right angles to the ray , determine the displacement of the point from it’s true position on the plan if scale if the plot was 1cm=1m.(Ans: 1mm)

78. The area within the contour lines at the site of reservoir and the face of a proposed dam are as follows :

Contours(m) Area(m²)300 620302 8 400304 60 240306 90 510308 100 200310 301 500312 70 300314 450 500316 527 280

Taking 300m as the bottom level of the reservoir and 314m as the water level , find the volume of water in the reservoir

(Ans: 1 713 420m3)

79. The contour interval on map is 12m . if the upward gradient of 1 in 20 is required to be drawn between two points, what will be the horizontal equivalent (sol: Horizontal equivalent = contour

interval × gradient

= 12×20=240m80. From a topographic map , the areas

enclosed by contour lines for a proposed dam are given below . find the volume of impounded water using trapezoidal formula. Contours (m) Arear enclosed (hectares) 500 20 505 100510 400515 900520 1100

81. Calculate the ordinates at 10m distance for a circular curve having a long chord of 80 m and a versed size of 43

(Ans:

o1=3.75 m, O2=3.01m ,O3=1.76 m)

82. In a town –planning scheme, a9m wide road is to intersect another road 12m wide

at 600, both being straight. The kerds

forming the obtuse angle by one of 120m radius .Calculate the distance required for setting out the four tangent pointsDescribe how to set out the larger curve by deflection angle method and tabulate the angles for 15m chords.

(Ans : 69.28,51.96m,δ=30 35 ')83. Two tangents PQ and QR to a railway

curve meet at an angle of 1400. Find the

radius of the curve which will pass through point M , 24m away from the intersection point Q, the angle PQM being

1000.

( Ans: 320.4m)84. In setting out a circular railway curve , it

is found that the curve must pass through a point 15m from the intersection point and equidistant from the tangents. The chainage of the beginning and end of the curve and the degree of the curve for a 20m chord

(ans: 490m,2757.83m,2997.17m,2.340)

85. A new railway line is to have its centre line on the courses of a traverse , the details of part of which are as follows:

course length BearingBC 1458.20 85029 ' 20 ' 'CD 180.20 9909 ' 00 ' 'DE 84016 ' 40 ' '

If minimum straight between curves is 20m, find the maximum allowable radius and the chainage of the four tangent points the chainage of B being 1381.51(ans: 640m,1381.54m,1534.14,1554.14m,1720.26m)

86. A compound curve is to connect of an arc of 900m radius followed by one of 1200m radius and is to connect two straight

intersecting at an angle of 93028 ' . At the

intersection point, the chainage, if continued alomg the first tangent, would

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be 2329.20m and the starting point of the curve is selected at chainage 1354.20m . calculate the chainage at the junction point of the two branches and at the end of the curve(ans: 1850.57m,3001.09m)

87. A right hand circular curve is to curve is to connect two straights PQ and Qr, the

bearings of which are 630 30 ' and

1200 42 ' Respectively. The curve is to

pass through a point S such that QS is

79.44m and the angle PQS is 34036 ' . Determinr the radius of the curve . if the chainage of the intersection point is 2049.20m, determine the tangential angles required to set out the first two pegs on the curve at through chainage of 20m

(Ans: 452.6m.0050 ' 8 ' ' ,10 15' 57 ' ' )88. A reverse curve is to be run from a point

T 1 on AA1 to the point T 2 on CC ' .

determine the common radius, and the lengths of the two parts of the curve, given

that T 1T is 720m and the angles AT 1T 2

and T 1T 2C ' are 470 30' and 250 12' , respectively .(Ans: R=301.496m,length of first arc=448.86m,length of second arc=331.51m)

89. Find the length of the vertical curve connecting two uniform grades from the following data.a) +0.8% and -06%;rate of change of

grade=0.1%per30mb) -0.5%and+1%;rate of change of

grade=0.05%per30m(Ans: 420m,p00m)

90. If the degree of a curve is 10 and if the

chain length is 30m, then the radius of the curve is equal toa) 5400mb) 1720m

c)1720

πm

d)3440

πm

91. The radius of a sample circular curve is 30m and the length of the sepecified chord os 30m. the degree of the curve osa) 57.29

b) 3.70c) 55.60d) 37.03

92. If the angle of intersection of curve is θ, then the defection angle will be

a)θ2

b) 1800−θc) 1800+θd) 900+θ

93. If S is the length of a sub-chrod and R is the radius of a simple curve , the deflection between its tangent and sub-chord, in mintutes, is equal toa) 573S/Rb) 1718.9S/Rc) 1718.9R/Sd) 573R/S

94. For a curve of radius 100m and normal chord 10m, the deflection angle given by Rankine formula is

a) 10 45.95 'b) 20 51.53 'c) 0035.9 'd) 10 25.95 '

95. If ∇ is the angle of deflection of the curve

and T 1 and T 2 are its points of

tangencies, the angle between the tangent

at T 1 and T 2 and along chord T 1T 2 will

be a) ∆/4b) ∆/3c) ∆/2d) ∆

96. Shift of a curve isa) L /6R²b) L/24Rc) L /24R²d) L /6R²97. The data of a closed traverse survey is shown below. determine the area

line Latitude(m) Departure(m)AB -300 +450BC +640 +110CD +100 -380DA -440 -180

(Ans: 253 100 m³)

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98. In order to obtain the area of a plot, series of perpendicular offsets 2.2m,3m,1.65,2.46,2.00m,2.25m and 1.68m were laid from a survey line to an irregular boundary at regular intervals of 5m. find the desired area using : (a) Trapezoidal rule (b) Simpson’s rule .(Ans: 66.5 m²,70.03m²)

99. Calculate the area enclosed by a traverse ABCD for the following data

line Latitude(m) Departure(m)AB +320.00 +40.20BC -3.00 +92.00CD -97.85 +6.402DA -15.64 -107.00EA +84.6 -31.60

(Ans: 387.44cm²)100. One hectare of an area is eqyivalent to

a) 10²m²

b) 104 m2

c) 106 m2

d) 109 m2

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