Sub: SurveyingTopic : 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 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] An area actually measures 0.8094 hectares. How much will it measure in 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 ] The area of apiece of a land which had been surveyed with a chain was calculated to be 9562 . Of this, 8935 was the total area of the triangles and 627 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 ] 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] The following are the observed fore bearings of the line. Calculate the respective back bearings. (i) AB (ii) CD (iii) EF (iv) GH (v) AB E (vi) CD E (vii) EF W (viii) GH [Ans: (i) AB (ii) CD (iii) EF (iv) GH (v) AB (vi) CD (vii) EF E (viii)GH Convert the following whole circle bearings to quadrantal bearings. (i) (ii) (iii) (iv) [Ans: , , , ] 6. 2. 3. 7. 4. A line measured with a steel tape which was exactly 30 m at a temperature of C and a pull of 10 kg. the measured length was 1650 m. the temperature during measurement was C 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 . The coefficient of thermal expansion of the material of the tape C is 3.5× and the modulus of elasticity of the material of tape is 2.1× kg/ . [Ans: 27.7768 m] 8. Convert the following quadrantal bearings to whole circle bearings. (i) E (ii) (iii) (iv) JH ACADEMY Page 1 Sub: Surveying [Ans: ] 14. Topic : Basics of surveying 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 W, find out the true bearings. Line Fore bearing Back bearing AB S E N W BC N E S W CD N W S W DA S W N E [Ans:AB=S E,BC=N E, CD= N W, DA= S ] 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 N S QR N S RS ST S TP [Ans: TPQ= ] 16. If the fore bearing of a line is zero degree, its back bearing is a) b) c) d) Which will be included angle AOB if the bearings of the lines AO and OB are respectively and ? a) b) c) d) Which will be the included angle AOB if the bearings of the lines AO and OB are respectively and ? a) b) c) d) If the magnetic bearing of a line is and the magnetic declination true bearing of the line will be a) b) c) d) 9. Find the value of magnetic declination if the magnetic bearing of the sun at noon is 25 ? [Ans: E] 10. 11. A line was drawn to a magnetic bearing of S5 0 E on old plane when the magnetic declination as 4 W. to what bearing should it be set now if the present declination is 2 E? [Ans: S5 E] The bearing of the sides of a triangle ABC are as follows. Computer the interior angles. AB = , BC = 13 , and CA= 27 [Ans: 15. 12. Find out the bearing of the lines of an equilateral triangle ABC running clockwise, if the bearing of the line AB is 6 3 . [Ans: BC = 18 30 3 3 , CA = 13. The following bearing were observed in running a compass traverse: Line F.B. B.B. AB 6 1 24 0 BC 12 4 31 3 CD 21 3 3 3 DA 30 4 12 4 Find the correct fore and back bearings and the true bearings o the lines, given that the magnetic declination is 4 E. [Ans: True bearing AB = 6 5 , BC = 13 4 , CD = 21 1 , DA = 30 2 ] 17. 18. 19. , the 20. The force bearing of line AB is . The included angle ABC is . The F.B. of the line BC is JH ACADEMY Page 2 Sub: Surveying a) 21. b) c) d) 25. Topic : Basics of surveying A helicopter files in sky from Manipuri(M) in Allahabad to Teliarganj (T) as per following conditions: a) 5.0 km along a up gradient eastwards up to B b) 3.0 km along an up gradient northwards up to C c) 4.0 km along an up gradient in the N-W direction up to D d) 4.0 km along a 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 E200 m and its height above datum is 100 m. [Ans: Bearing of TM= The magnetic bearing of the sun at noon is . The magnetic declination at the place is a) b) c) d) S The bearings to two inaccessible stations A and B taken from a station C were and , respectively. The coordinates of A and B were as follows: Station Easting Northing A 300 200 B 400 150 Calculate the independent coordinates of C. [Ans: 363.45, 223.10] 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 335 BC CD 408 DA 828 Calculate the missing bearing and the length. [Ans: Bearing = , Length = 849.5 m] 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 AD 731 DE 467 EF 583 Calculate the necessary length and bearing of PC. [Ans: length= 492.85 m, Bearing = ] 26. 22. ] 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 EF Northi 590 606 101 ng 1190. 565 .5 .9 7.2 9 .3 eastin 0 736 796 370. g .4 .8 46 4 8.0 If the bearing of FG is and its length is 896.00m, find the length and bearing of GA. [Ans: 884.21 m, For a railway project, a tunnel is to be run between two points P and Q points coordinates N E P 0 0 Q 4020 800 R 2120 1900 It is desired to sink a shaft at S, the midpoint 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; ] 23. 24. 27. JH ACADEMY Page 3 Sub: Surveying 28. A surveyor is running a line in the direction AB having a bearing N 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 BC 120.00 C (L) CD 280.00 D [Ans: 28.48 m, ] For a closed ABCDA, missing data. Line Length (m) AB 100.00 BC 605.00 CD 95.00 DA ? [Ans: 679.97 m; 30. compute the bearing N N N ? ] Topic : Basics of surveying [Ans: 10 4 ,DE = 1776.89 m, CD = 1170.96 m] 32. A straight road AB is proposed to be extended in the direction AB produced, the / 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 = 8 BD = 29.02 m BDE = 28 DE = 77.14 m DEC = 29 Calculate a) the length of lime EC. b) the angle to be measured at C, so that the c/ 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, ] 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] The following staff readings were taken with a level. The instrument having been shifted after the 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] 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 29. In a quadrilateral ABCD, the coordinates of the points (in metres) are as follows: point E N A 100 100 B 208 104 C 223 353 D 100 357 Calculate the area. [Ans: 29221.5 ] The following observations were made for a closed transverse ABCDEA. Line Length (m) Included angles AB 1512.1 BC 863.7 CD ? 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. 34. 31. 35. JH ACADEMY Page 4 Sub: Surveying 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) 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 Height of Staff Staff at instrument at 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)] 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 m a) 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. Topic : Basics of surveying [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] 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] 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] 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 Remarks 2.405 Point A(RL=250.05 m) 1.954 1.128 C.P 0.169 1.466 Point B 2.408 Point D -1.515 Point E 1.460 2.941 C.P 2.368 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] A line of levels was run in the form of a loop 650 m long. The initial starting point 40. 41. 37. 42. 38. 43. JH ACADEMY Page 5 Sub: Surveying 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 A 0.445 2.370 2.150 0.995 0.72 B 1.465 0.260 2.630 0.905 1.305 40.0 B.M (ANS: R.L of A =37.950M ,R.L. of B=97.655 M) In a two-peg test of a dumpy level, the following readings were taken : Instrument at R Staff readings Midway between at P 1.525m pegs P and Q at Q 1.230m Instrument near P at P 1.420m At Q 1.025m How 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] Two pegs A and B driven 100.0m apart. In the adjustment of a dumpy level, the following observations were recorded Topic : Basics of surveying 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] 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 and , respectively. Angle of elevation as observed at P and Q to the top F were and 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) Find the reduced level of the top of a church apire from the following date Instrument station Reading on B.M Vertical angle Remark 48. 49. 45. A 50. R.L of B.M : 543.075m B 1.269 Distance AB =30m Stations A and B the church spire are in the same vertical plane. (Ans: 566.628) To determine the elevation of the top of a chimney the following observations were made: Instrument station Reading on B.M Vertical angle Remark 1.578 46. P 1.377m R.L of B.M : 50.15m INSTRUMENT NEAR STAFF READINGS on A B R.L(m) A 1.520 1.535 80.00 B 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 30’. The horizontal distance between the instrument 51. Q 1.263m 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) 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 and at B as . 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 JH ACADEMY Page 6 Sub: Surveying found to be . Calculate the elevation of the tower (Ans: 256.284m) 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) 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: ) 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 and that observed from instrument station B is . the angle of elevation from B to a vane 1.75m above the foot of the staff held at station A is . 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) 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 and 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) The following observations were made in running fly levels from a bench mark of R.L. 60.65: Back sight : 0.964,1.632,1.105,0.850 Topic : Basics of surveying Fore sight 0.948,1.153,1.984 Five 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. (Ans: staff reading of peg 1:1.116m peg2:1.316m, peg3:1.516m,peg4:1.716m,peg5:1.916m) Correction due to refraction is given by a) 0.0112D² b) 0.0785D² c) 0.0673D² d) 0.0012D² What will be the correction for curvature for a distance of 1000m? a) 0.0673m b) 0.0785m c) 78.50m d) 6.73m 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 is a) 101.250 b) 102.0 c) 104.50 d) 103.250 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) 2.8938m b) 2.8150m c) 2.8961m d) 2.8950 m 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) 0.015m b) 0.30m c) 0.045m d) 0.060m Following observations were recorded with a tachometer fitted with an analecticlens (K=100,C=0). Calculate the 52. 53. 57. 58. 54. 59. 60. 55. 61. 56. 62. JH ACADEMY Page 7 Sub: Surveying 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 O O H.I 1.500 1.500 Staff station B.M C.P Vertical angle Staffs readings 1.250,1.400,1.550 1.550,1.750,1.950 Topic : Basics of surveying (Ans: PQ=422.13m, R.L=335.46m) 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 Instrumen t station Staff statio n B.M Vertica l angle Staff readings(m) Remarks 66. T 1.350 C.P 1.390,1.550,1.710 1.100,1.153,2.06 0 63. (Ans: 506.291 m, 510.767 m) 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) R.L of B.M =976.00 m A B 0.982,1.105,1.18 8 67. top A middle bottom C 1.044 2.283 3.522 D 0.645 2.376 4.100 The line BA has a bearing of 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. 64. ) 68. (ans: R.L=988.81m) 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 and in the second case is . 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) 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 O Staff station P Q bearing Vertical angle Staffs reading(m) 1.365,1.920,2.475 1.065,1.885,2.705 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 A B C 1.140,2.292,3.420 Horizontal angle Vertical angle Staff readings(m) 69. 65. 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) 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) 2.250 2.055 Axial hair readings (m) 2.105 1.875 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 instru ment H.I( m) Total coordinates Staf f stati on A Q.B Vert ical angl e Staff reading( m) 1.65,2.7 5,3.85 C 1.4 50 N 300.0 0 E 812.1 70 N D 1.5 00 586.6 500 1250. 750 B 2.50,3.2 0,3.90 Calculate : i) Length of line AB ii) Gradient of line A/b iii) Reduced levels of A and B (Ans:673.83m,1in170.16(rising),1158.29 m,162.24m) 70. The constant for an instrument are 1000 and 0.5. calculate the distance from the P Q 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 JH ACADEMY Page 8 Sub: Surveying 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 , the staff being held vertical (Ans: 189.95 m) 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 + and . Find the horizontal distance and R.L of A, if the height of the line of collimation is 100m (Ans: 42.3m ,98.388m) 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 and , respectively. If the lower targets was at an elevation of 249.2m, what was the height of instrument (Ans: 240.43m) The multiplaying constant of a tacheometer is a) f/i b) (f/d)+i c) (f/i)+d d) F+d For a tacheometer equipped with an anallatic lens , the additive and multiplying constants are, respectively a) 0 100 b) 100 0 c) 0 0 d) 100 100 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 is a) 0.1 b) 0.4 c) 1.66 d) 20 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 a) 0.75m b) 7.5m c) 75m Topic : Basics of surveying d) 750m 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) 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) 300 620 302 8 400 304 60 240 306 90 510 308 100 200 310 301 500 312 70 300 314 450 500 316 527 280 Taking 300m as the bottom level of the reservoir and 314m as the water level , find the volume of water in the reservoir (Ans: 1 713 420 ) 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=240m 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 20 505 100 510 400 515 900 520 1100 Calculate the ordinates at 10m distance for a circular curve having a long chord of 80 m and a versed size of 43 (Ans: ) In a town –planning scheme, a9m wide road is to intersect another road 12m wide at , both being straight. The kerds 77. 71. 78. 72. 73. 79. 74. 80. 75. 76. 81. 82. JH ACADEMY Page 9 Sub: Surveying forming the obtuse angle by one of 120m radius . Calculate the distance required for setting out the four tangent points Describe how to set out the larger curve by deflection angle method and tabulate the angles for 15m chords. (Ans : 69.28,51.96m,δ= ) Two tangents PQ and QR to a railway curve meet at an angle of . Find the radius of the curve which will pass through point M , 24m away from the intersection point Q, the angle PQM being . ( Ans: 320.4m) 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, ) 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 Bearing BC 1458.20 CD 180.20 DE 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) 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 . At the intersection point, the chainage, if continued alomg the first tangent, would 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) A right hand circular curve is to curve is to connect two straights PQ and Qr, the bearings of which are and Topic : Basics of surveying Respectively. The curve is to pass through a point S such that QS is 79.44m and the angle PQS is . 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. , ) A reverse curve is to be run from a point on to the point on . determine the common radius, and the lengths of the two parts of the curve, given that is 720m and the angles and are and , respectively . (Ans: R=301.496m,length of first arc=448.86m,length of second arc=331.51m) 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%per30m b) -0.5%and+1%;rate of change of grade=0.05%per30m (Ans: 420m,p00m) If the degree of a curve is and if the chain length is 30m, then the radius of the curve is equal to a) 5400m b) 1720m c) d) 91. m 83. 88. 84. 89. 85. 90. 86. 92. The radius of a sample circular curve is 30m and the length of the sepecified chord os 30m. the degree of the curve os a) 57.29 b) 3.70 c) 55.60 d) 37.03 If the angle of intersection of curve is θ, then the defection angle will be a) b) θ c) θ d) If S is the length of a sub-chrod and R is the radius of a simple curve , the deflection between its tangent and subchord, in mintutes, is equal to 87. 93. JH ACADEMY Page 10 Sub: Surveying a) 573S/R b) 1718.9S/R c) 1718.9R/S d) 573R/S For a curve of radius 100m and normal chord 10m, the deflection angle given by Rankine formula is a) b) c) d) If is the angle of deflection of the curve and and are its points of tangencies, the angle between the tangent at and and along chord will be a) ∆/4 b) ∆/3 c) ∆/2 d) ∆ Shift of a curve is a) L²/6R b) L/24R c) L²/24R d) L²/6R The data of a closed traverse survey is shown below. determine the area line Latitude(m) Departure(m) AB -300 +450 BC +640 +110 CD +100 -380 DA -440 -180 (Ans: 253 100 m³) d) Topic : Basics of surveying 94. 95. 96. 97. 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 +40.20 BC -3.00 +92.00 CD -97.85 +6.402 DA -15.64 -107.00 EA +84.6 -31.60 (Ans: 387.44cm²) 100. One hectare of an area is eqyivalent to a) 10²m² b) c) 98. JH ACADEMY Page 11