UNIT-1 Short Answer Type Questions
UNIT-1Short Answer Type Questions:1. State the Objectives of Surveying?2. What is basic principle on which Surveying has been classified? And explain them?3. Differentiate between Plane Surveying & Geodetic Surveying?4. State the various types of functional classification of Surveying?5. State the various types of Surveys based on the instrument used?6. Explain the term ‘Cartography’?7. Explain the Principle ‘Always work from the whole to the part’, with an example.8. What are the Basic Measurements in Surveying & state the names of the instruments used tomeasure these basic Measurements9. What are the units of measurement for ‘Linear Measurements’10. What are the different systems of units of measurement for ‘Angular Measurement’11. Explain Plan and a Map? And differentiate them?12. Explain ‘Scale of a Map’ and how the scale has been classified and its range of values?13. Explain the term ‘Representative Factor’14. State the major classification of ‘Duties of a Surveor’15. Define the terms Accuracy and Precision with the help of an example?16. What are the Sources of errors? And explain each with 3 to 5 bullet points17. What are the Types of errors? And explain each with 3 to 5 bullet points18. Explain the following approximate methodsa. Pacingb. Passometerc. Pedometerd. Odometere. Measuring Wheelf. Speedometer19. Explain the Parts of the Chain with the help of a neat sketch.20. Explain the process of Folding and Unfolding of the chain21. Compare the Advantages and Disadvantages of Tape over Chain.22. Explain the following instruments with 2 to 4 point bullets each
a.b.c.d.e.f.g.ArrowsPegsRanging rods & Ranging polesOffset RodsPlumb bobSimple ClinometerPlasterer’s Laths & Whites23. Explain the terms ‘Ranging Out’ and State the different methods of Ranging out.24. What are the Code of Signals used during Surveying and it’s meaning. Explain with the help ofTable and Sketch.25. What are Chain Corrections and state the formulaes.26. State the various types of Tape Corrections27. State the possible errors in Chaining?28. Explain the Principal of Chai Surveying29. Explain the following termsa.b.c.d.e.f.g.h.i.Main Survey StationsSurvey LinesCheck LinesOffsetsRange TiesPlus MeasurementsTie LinesSubsidiary StationsBase Line30. Explain ‘Well Conditioned Triangle’31. Explain ‘Field Book’32. Explain ‘Optical Square’33. Explain any 2 methods of Chaining across Obstacles34. Explain any 2 methods of Ranging across Obstacles35. What are the different types of conventional symbols used to denotes various objects andtabulate them36. What is meant by ‘Traverse’? Explain its types?37. Difference between Traverse Surveying’ and Chain Surveying38. What is a Meridian? And explain its types?
39. What is a bearing? And explain its types?40. Explain ‘Designation of Bearings’ in detail41. Explain Fore Bearing & Back Bearing42. Comparison between Surveyor’s Compass & Prismatic compass?43. What are the temporary adjustments to be done for Surveyor’s Compass & Prismatic compass?44. State the Permanent Adjustments of a Compass?45. Explain the term ‘Magnetic Declination’46. What the different types of ‘Magnetic Declination’ and explain each by 2 bullet points47. Explain True bearing and Magnetic Bearing?48. Explain the term ‘Local Attraction’ in about 6 to 8 bullet points49. What are the errors in compass surveying50. Explain the term ‘Dip’
Long Answer Type Questions:1. Explain the following termsa.b.c.d.e.f.g.h.i.j.k.l.m.n.o.p.Control SurveyingLand SurveyingCity SurveysTopographical SurveysRoute SurveyingMine SurveysHydrographic SurveysEngineering SurveysAstronomic SurveysSatellite SurveysGeological SurveysConstruction SurveysArchaeological SurveysMilitary SurveysGravity SurveysGeneral Surveys2. Explain the following terms (each in 2 or 3 bullet points)a.b.c.d.e.f.g.h.Chain SurveyingCompass SurveyingLevellingPlane Table SurveyingTheodolite SurveysTacheometric SurveysPhotogrammetric SurveysEDM Surveys3. State the important Principles of Surveying. And explain each term with 2 or 3 bullet points.4. Explain the following with 5 bullets points eacha. Works involved in the Field Work of the Surveyorb. Works involved in the Office Work of the Surveyor5. Care and Adjustments of Instruments6. Explain the followinga. Correction for Standardizationb. Correction for Slopec. Correction for pulld. Correction for temperaturee. Correction for sagf. Correction for misalignment
7. Explain the followinga. Direct Rangingb. Indirect Rangingc. Random line Method8. What are the different types of Tape? And explain each type with 3 bullet points9. What are the different types of Chains? And explain each type with 3 bullet points
Problems:1. A rectangular plot of area 14144 sqm is represented by 8.5cm x 6.5cm on map. Find out thescale of map & mention its R.F.2. A rectangular plot of land of area 0.45 hectare is represented on a map by a similar rectanglearea of 5 cm2. Calculate the R.F of the scale of the map.3. A plan represents an area of 93,750 m2 and measures 6 cm x 6.25 cm. Find the scale of the plot.4. A 1.2 Km long road is indicated in a map by a length of 30cm. Find the scale of the plot.5. The Plan of an area has shrunk such that a line originally 10cm now measured 9.5cm. If theoriginal scale of the plan was 1cm 10m (R.F. 1:1000), determine (i) the shrinkage factor, (ii)Shrunk Scale, (iii) Correct distance corresponding to a measured distance of 98m, (iv) Correctarea corresponding to a measured area of 10,000m2.6. A rectangular plot in pln is 10cm x 30cm, drawn to a scale of 1cm 100m. If the same plot isredrawn on a toposheet to a scale of 1cm 1km, what should be its area on the toposheet?Determine also R.F in each case.7. Determine the scale of plan and the representative fraction in the following two casesa. A line 150m long is represented by 7.5cm on plan.b. A ground area of 112.5m2 is represented by 4.5cm2 on plan.8. If a line originally 5cm has shrunk to 4.5cm, determine the shrinkage factor. Also determine thecorrect area corresponding to a measured area of 400m2.9. Find the max. Permissible error in laying off the direction of offset so that the max. displacementmay not exceed 0.25mm on the paper, given that the length of the offset os 10m, the scale is20m to 1 cm and the max. error in the length of the offset is 0.3m.10. A Surveyor measured the distance between 2 points marked on the plan drawn to a scale of 1cm 1m (R.F 1:100) and found it to be 50 m. Later he detected that he used a wrong scale of1cm 50 cm (RF 1:50) for the measurement.a. Determine the correct lengthb. What would be the correct area if the measured area is 60m211. An angle was measured six times, the observed values being 49023’00”, 49023’20”, 49022’40”,49022’20”, 49023’40”, 49024’00”. Calculate the most probable value of the angle and thestandard error of the measurement.12. The measurements taken by 2 survey parties for the same line were as follows:a. First Party: 50.385m, 50.394m, 50.380m and 50.373mb. Second Party: 50.364m, 50.365m, 50.363m & 50.364mIF the true value of the line is 50.382m, state which measurements are accurate & which areprecise.
13. A 30m chain was found to be 3cm too long after chaining 1800m. It was 9cm too long at the endof day’s work after chaining a total distance of 300m. If the chain was correct beforecommencement of the work, find the time distance.14. The length of a survey line when measured with a chain of 20m nominal length was found to be841.5m. When the chain was compared with a standard, it was found to be 0.1m too long. Findthe correct length of the line.15. The length of a survey line measured with a 30m chain and was found to be 315.4m. When thechain was compared with a standard, it was found to be 0.2m too short. Find the correct lengthof the line.16. The distance between 2 points measured along a slope is 518m. Find the horizontal distanceb/w them if (a) the slope angle if 6 degrees. (b) the slope is 1 in 5.17. If the ground rises by 5m in a chain length of 20m, determine the slope correction. Compare theresult with that obtained by approximate formula.18. A steel tape 30m long was standardized with a pull of 65N. If the pull at the time ofmeasurement was 45N, find the correction per tape length. The tape weighs 10N. Take E 2 X105 N/mm2 and weigh of 1 m3 of steel as 77.10 KN.19. The measured length of a line was 850.42m, with a 30m tape which was exactly 30m at 27 0C.Find the correct length. Take a 1.16 x 10-5 per 0C. The field temperature was 320C.20. Determine the sag correction for a 30m steel tape under a pull of 80N in 3 bays of 10m each.The area of the cross-section of the tape is 8mm2 and the unit weight of steel may be taken as77 KN/m2.21. Determine the normal tension for a steel tape supported b/w two supports 10m apart if thestandard tension is 65N and the weight of tape per metre is 0.62N. Take E 2 X 105 N/mm2 andthe area of Cross section as 8 mm2.22. A survey line was measured to be 60m. It was found that there was misalignment and the linewas 1m off the straight line at the middle. Determine the correct length.23. A distance was measured using a 30m steel tape in 4 sections: 30m, 30m, 30m, 26.455m, a totalof 116.455m, The tape was supported at 2 ends during the measurements. The fieldtemperature was 300C and a tension of 100N was used. The tape was calibrated fully supportedat a temperature of 200C using a tension of 75N and had a length of 30.01m. Compute thecorrect distance. Take the weight of the tape as 15N and c-s area as 0.02 cm2. Co-eff ofexpansion 1.16 x 10-5/0C, Young’s modulus E 2.06 x 103 N/mm2.24. During a survey, a time was measured with a steel tape which was exactly 30mts at 20 0C at apull of 100N, the measured length being 1500 m. The temperature during measurement was300C and the pull applied was 150N. The cross sectional area of the tape was 0.025 Sqcm. The
co-efficient of expansion of the material of the tape per 10C 3.5 x 10-6 and the modulus ofelasticity of the material of the tape 2.1 x 10-5 N/mm5. Find the length of the line after applyingcorrections.25. A steel tape of nominal length 30m was used to measure a line AB by suspending it b/wsupports. If the measured length was 29.861m when the slope angle was 3045’, and the meantemperature and tension applied were respectively 10 C and 100 N, determine the correcthorizontal length. The Std length of the tape was 30.004m at 200 C and 44.5N tension. The tapeweighed 0.16 N and had a cross-sectional area of 2mm2, E 2 X 105 N/mm2, α 1.12 x 10-5 per C.26. Convert the following whole circle bearing to quandrantal bearing:a. 22030’b. 170012’c. 211054’d. 327024’27. Convert the following quadrantal bearing to whole circle bearingsa. N 22030’Eb. S31030’Ec. S6806’Wd. N5042’W28. The following are observed fore-bearings of the linesa. AB 12024’b. BC 112048’c. CD 266030’d. PQ S59018’We. QR N1800’E29. The Magnetic Bearing of a line AB is 88045’. Calculate the true bearing if (a) Magneticdeclination is 5030’ east, (b) the magnetic declination is 4045’W.30. The magnetic bearing of a line AB is S38030’W. Calculate the true bearing if the magneticdeclination is (a) 4030’W (b) 3030’ E.31. A line AB was drawn to have a magnetic bearing of 25030’ in an old map when the declinationare 2030’E. Determine the magnetic bearing of the line if the present declination is 5030’W.32. A line AB had the magnetic bearing of 44030’ in 1910 when the declination was 4030’W.Determine the magnetic bearing of the same line in 1990 if the annual change was 6’ eastwards.33. The Following bearings were observed for a closed traverse ABCDEA.Line AB – 140030’Line BC – 80030’Line CD – 34000’Line DE – 290030’Line EA - 230030’
Calculate the included angles.34. In a closed traverse ABCDE, the bearings of the line AB was measured 150030’. The includedangles were measured as under: A 130010’, B 89045’, C 125022’, D 135034’, E 5909’Calculate the bearings of all other lines.35. Following is the data regarding a closed compass traverse PQRS taken in clockwise direction.a. Fore bearing & Back bearing at Station, P 550 & 1350b. Fore bearing & Back bearing of line, RS 2110 & 310c. Included angles Q 1000 & R 1050d. Local Attraction at station R 20 We. All observations were free from all the except local attractionFrom the above data:i. Calculate the local attraction at station P & Sii. Calculate the corrected bearing of all the lines and tabulate the same.36. In a closed traverse ABCDE, the bearings of the line AB was measured as 150030’.The includedangles were measured as under.Angle A 130010’, Angle B 89045’, Angle C 125022’, Angle D 135034’, Angle E 5909’Calculate the bearings of all other lines.37. A Compass traverse ABCDEA, was run anticlockwise and the following bearings were takenwhere local attraction was suspected.LineFBBB0AB150 DE314015’134045’EA220015’40015’Determine the local attraction and the correct bearings by the both the methods.38. A Compass traverse ABCDEA, was run anticlockwise and the following bearings were takenwhere local attraction was suspected.Determine the correct included ��40015’
39. Given below are the bearings observed in a traverse survey conducted with a prismatic compassat a place where local attraction was suspected.LineFBBB0AB124 ’DA200015’17045’At what stations do you suspect local attraction? Find the correct bearings of the lines and theincluded angles.40. The following bearings were taken in running a closed traverse while surveying in Jhansi, AllahabadLineFBBB0AB48 5’DE165015’345015’EA259030’79000’ State the stations which are affected by local attraction and by how much. Determine the correct bearings. Calculate the true bearings, if the declination was 1030’W
OBJECTIVE TYPE QUESTIONS:1. The curvature of the earth’s surface, is taken into account only if the extent ofsurvey is more thani) 100 Sq.kmii) 160 Sq.kmiii) 200 Sq.kmiv) 250 Sq.km2. The difference in the lengths of an arc and its subtended chord on the earth’s surfacefor a distance of 18-20 km is onlyi) 1 cmii) 5 cmiii) 10 cmiv) 1003. In geodectic surveys, higher accuracy is achieved if(i) Curvature of the earth surface is ignored(ii) Curvature of earth surface is taken into account(iii) Angles between the curved lines are treated as plane angles(iv) None of the above4. Hydrographic surveys deal with the mapping ofi) Large water bodies ii) Heavenly bodies iii) Mountainous region iv) movementof roads5. Surveys which are carried out to depict mountains, rivers, water bodies, wooded areasand other cultural details are known asi) Cadastral surveysii) City surveys iii) Topographical surveysiv) Guidemap surveys v) Plane surveys6. Surveys which are carried out to provide a national grid of control for preparationof accurate maps of large areas, are known asi) Plane surveysii) Geodetic surveysiii) Geographical surveysiv) Topographical surveys7. The main principle of surveying is to worki) From part to the wholeii) from whole to partiii) From higher level to lowerlevel iv) From lower level to higher level8. EDM instruments are used for measurement ofi) Short distances ii) medium distancesiii) Long distances iv) None of the above9. Remote sensing is defined as the collection of information about a subjecti) with direct physical contact ii) without direct physical contact iii) either i) or ii)iv) None of the above10. In the D120 distomat instrument, by touching a selector key, distances can be displayedin metres or feet. The least count isi) 1 mmii) 2 mmiii) 3 mmiv) 4 mm11. . Short offsets are measured withi) An ordinary chainii) An invar tapeiii) A metallic tapeiv) A steel tape
12. . Greater accuracy in linear measurements is obtained byi) Tacheometryii) Direct chainingiii) Direct tapingiv) All the above13. In which chain, each metre length is divided into 5 links?i) 20 mii) 30 miii) Both iv) None of the above14. While measuring a line between two stations A and B intervened by a raised ground,i) Vision gets obstructedii) Chaining gets obstructediii) Vision and chaining get obstructediv) None of the above.15. Accurate measurement of distance are made withi) Chainii) Invar tapeiii) Metallic tapeiv) Steel Tape16. In chain surveying a tie line is primarily provide(i) To check the accuracy of the survey(ii) To take offsets for detailed survey(iii) To avoid long offsets from chain lines(iv) To increase the no. of chain lines17. In chain surveying fieldwork is limited to(i) Linear measurements only(ii) Angular measurements only(iii) Both linear and angular measurements(iv) All the above18. Check lines in chain surveying are essentially required(i) To plot chain lines(ii) To plot offsets(iii) To indicate the correctness of the survey work(iv) To increase the turn out19. A well conditioned triangle does not have any angle less thani) 20 ii) 30 iii) 45 iv) 60 20. Chain surveying is well adopted fori)Small areas in open groundii)Small areas with crowded detailsiii)Large areas with simple detailsiv)Large areas with difficult details21. The prismatic compass readsi) WCB ii) RB iii) Quadrantal bearing iv) None of the above22. Fore and back bearings of a line should differ byi) 90 ii) 180 iii) 360 iv) None of the above
23. The sum of the interior angles of a closed traverse is equal toi)(2n-4) x 90 ii) (2n 4) x 90 iii) (n-4) x 90 iv) None24. A negative declination shows that the magnetic meridian is to thei) West of the true meridianii) East of true meridianiii) South of true meridianiv) North of true meridian25. In the Quadrantal bearing (Q.B.) system, a line is said to be free from local attraction, ifthe forward and backward bearing arei) Numerically equal with same quadrantii) Numerically equal with opposite quadrantiii) Difference is 90 algebraicallyiv) The WCB’s are same.26. In the WCB (whole circle bearing) system, a line is said to be free from local attraction ifthe difference between Fore bearing and Back bearing isi)270 ii) 90 iii) 0 iv) 180 27. The least count of the prismatic compass isi)15’ ii) 30’ iii) 45’ iv) None of the above28. The magnetic bearing of a survey line at any placei) Remains constant ii) Changes systematically iii) Varies differently in differentmonths of the year iv) Always greater than true bearing29. The magnetic meridian at any point, is the direction indicated by a freely suspendedi) Magnetic needle ii) Properly balanced magnetic needle iii) properly balancedand uninfluenced by local attraction iv) Magnetic needle overan iron pivotAnswers for Objective type 11011632222835311317123129362121182241
UNIT-IIShort Answer Type Questions:1. Define the term ‘Levelling’2. Define the following termsa. Vertical Lineb. Level Surfacec. Mean Sea Leveld. Geoid3. Explain the for basic terms in levellinga. Level lineb. Datum Surface or Datumc. Elevationd. Altitudee. Difference of Elevationf. Reduced level (R.L)g. Horizontal planeh. Horizontal Linei. Vertical Planej. Vertical Angle4. What is Bench Mark (B.M) and Explain the types of Bench Marks with 2 or 3 bullet points each?5. What is Dumpy level and explain its parts6. What are the temporary adjustments of a Dumpy level and explain them7. What is a levelling Staff and what are the different types of levelling Staff?8. Explain Basic Principles of Levelling?9. What is meant by Level Field Book10. Explain Theory of Simple Levelling11. Compare Height of Instrument Method and Rise & Fall Method12. What are the hand Signals used in Levelling? Explain them with neat sketch13. What is a Contour and Contour Interval?14. What the factors influencing for selecting the Contour interval? Explain each with 2 or 3 bulletpoints?
15. What the different methods of locating Contours?16. Write a short notes on Interpolation of Contours17. Write a short notes on Sketching of Contours18. What is Contour Gradient?19. Explain the terms Effect of Curvature and Effect of Refraction and state the formulaes?20. What are the errors in levelling?21. What are the Mistakes in levelling?22. List out the formulae’s for following methods for finding the area with straight irregularboundaries and explain them with the neat sketches.a. Mid Ordinate ruleb. Average Ordinate rulec. Trapezoidal Ruled. Simpon’s Rule23. Explain the following termsa. Meridian Distanceb. Double Meridian Distancec. Latituded. Departure24. What are the approximate methods of finding Areas?25. What is a Planimeter?26. Explain the parts of the Planimeter with the help of a sketch27. What is a Zero Circle?28. Explain One Level Section with neat sketch?29. Explain Two Level Section with neat sketch?30. Write the formula for Trapezoidal and Prismoidal formula for finding Volumes31. State the formula for Calculating Prismoidal correction for one level and 2 level sections32. What is meant by ‘Mass-Haul Diagram’?
LONG ANSWER TYPE QUESTIONS:1. Explain the following types of direct levellinga. Simple Levellingb. Differential Levellingc. Check Levellingd. Fly Levellinge. Profile Levellingf. Cross-section Levellingg. Reciprocal levellingh. Precise levelling2. Explain the following termsa. Stationb. Height of Instrument (H.I)c. Back sight (B.S)d. Fore Sight (F.S)e. Intermediate Sight (I.S)f. Turning Point or Change Pointg. Balancing of Sights3. Explain the following terms in detail with the help of a sketcha. Profile Levelling or Longitudinal Levellingb. Cross-sectioning4. Explain the Direct Method Contouring?5. What the Characteristics of Contours?6. Explain the following terms of Indirect Method of Contouring?a. Grid Methodb. Cross-section Methodc. Radial Lines Methodd. Controlling Point Method7. What are the Uses of Contours? And explain each with 3 or 5 bullet points and Sketches8. Differentiate the following with the help of Sketches (Contours)a. Hill – Depressionb. Ridge Line – Valley Linec. Gentle Slope – Steep Slope9. Draw neat contours maps showing,a) Pond with almost uniform slope in all direction (show 5 lines).b) Vertical cliff with lowest R.L 619m and highest R.L 620.5m at 0.5m interval.
c) Ridge line with minimum and maximum R.L values of 410m & 418m(show 5 lines).d) Valley time for 5 contours.10. From the following figure,answer the following questions,a. Identify the topographical feature.b. Roughly indicate the direction of steepest slope and most gentle slope.c. What will the approximate R.L of point P and value of R.L 1
PROBLEMS:SHORT TYPE:1. The Staff reading at a point A was observed as 2.920m. If the staff was 10cm of the verticalthrough its bottom, find the correct reading.2. The Staff reading with a 4m staff, at a point A was 3.50m, the top of the staff was found to be10cm off the vertical through the bottom of the staff. If the staff was held vertically, determinethe correct reading.3. A dumpy level was set-up mid-way b/w two peg points 80m apart. The readings on the staff atthe 2 pegs were 3.2m and 3.015m respectively. The instrument was then moved, by 20m aheadof the second peg, in line with the two pegs. The respective staff readings were 2.825m and2.69m. Calculate the staff readings on the 2 pegs to provide a horizontal line of sight.4. The following observations were taken during the testing of a dumpy level:Instrument atStaff readings atABA1.2752.005B1.0401.660If the instrument in adjustment? To what reading should the line of collimation be adjustedwhen the instrument is at B. If A & B were 100m apart, what the angle of inclination of the lineof collimation?5. Find the height of a Tee-beam above the floor level. The R.L of the floor is 100.855m and thestaff reading on the floor is 2.055m. The reading on a staff held upside down against theunderside of the beam is 3.565m.6. The reduced level of a factory floor is 30m and the staff reading on the floor is 1.40m. The staffreading when held inverted with the bottom touching the Tee-beam of the roof is 3.67m. Findthe height of the beam above the floor.7. Calculate the combined correction for curvature and refraction for a distance of (i) 5Km, (ii)500m.8. An observer stands on the top of a tower with his eye level at 85m. Determine the distance tovisible horizon and the dip of the horizon. The radius if the earth may be taken as 6400Kms.9. The captain of a ship just sees the top of a light house 75m in height. If the eye level of captainwas 7m above the sea level, determine the distance of the ship from the light house.10. A line of level was run in the form of a loop 600m long. The initial elevation of the starting pointwas 29m. When the last foresight reading of 3.005m was made on the starting point, the heightof instrument was 32m. Compute the closing error. Is it tolerable?
11. Find the combined correction for curvature and refraction for a distance of 3km.12. In order to find the difference in Elevation between two points A & B,a level was setupon the line AB, 50m from A and 1300m from B. A & B being on same side of theinstrument ,the readings obtained on staff held at A & B were 0.435m and 3.95mrespectively,Find the true difference in elevation between A & B.13. A line of level was run in the form of a loop 600m long.The initial elevation of thestarting point was 29.0m,when the last pre-sight reading of 3.005m was made on thestarting point,the height of instrument was 32.0m,Compute the closing error.Is ittolerable?14. A bubble tube has a sensitiveness of 23 seconds for a 2mm division.Find the error in staffreading at a distance of 100m caused by the bubble out by one division.15. Calculate the area of a plan from the following readings of a planimeterInitial reading 8.348, Final Reading 1.435The zero of the disc passed the fixed index mark twice in the clockwise direction. The anchorpoint was placed outside the plan, and the tracing point was moved in the clockwise directionTake M 100cm216. Find the area of the zero circle from the following observations. Take M 100 cm 2.a. Anchor point outside the figurei. Initial reading 8.436ii. Final reading 4.325The zero of the disc passes the fixed index marks once in the clock wise direction.b. Anchor point inside the figurei. Initial reading 2.844ii. Final reading 5.434The zero of the disc passed the fixed index mark twice in the anti-clockwise direction.17. The following notes refer to the reciprocal levels taken with one level:Instrument StationABStaff readings onA1.030.95RemarksB1.631.54Distance AB 800mR.L of A 450m18. The following notes refer to the reciprocal levels taken with one level:Instrument StationABFind:(i)(ii)Staff readings onA1.4251.429True R.L of B.Correction for CollimationRemarksB2.7242.504Distance AB 1150mRL of A 100m,Collimation error 0.003/150m
(iii)Correction for refraction(iv)Combined correction for Curvature and refraction.LONG TYPE:1. In order to find the difference in elevation between 2 points A & B, a level was set up theline AB, 50m from A & 1500m from B. A & B being on same side of the instrument. Thereadings obtained on staff held at A & B were 0.435m and 3.95m respectively. Find thetrue difference in elevation between A & B.2. The following readings have been taken from the page of an old level book. Re-constructthe page. Fill up the missing quantities and apply the usual 20573.6258x322.590B.M11. The following consecutive readings were taken with a level and 3m leveling staff on acontinuously sloping ground at a common interval of 20m: 0.602, 1.234, 1.860, 2.574, 0.238,0.914, 1.936, 2.872, 0.568, 1.824, 2.722, R.L of first point was 192.122. Calculate R.L of pointsand gradient of line joining first and last points.12. The following consecutive readings were taken with a leveling instrument at intervals of 20m:2.375, 1.730, 0.615, 3.450, 3.450, 2.835, 2.070, 1.835, 0.985, 0.435, 1.630, 2.255, 3.63m.Theinstrument was shifted after fourth and eighth readings. The last reading was taken on BM ofR.L 110.2m. Find the R.L’s of all the points.13. Reciprocal leveling between 2 points A & B, 630.5m apart on opposite sides of a river gave thefollowing results:Instrument AtABHI1.36m1.335mStaff atBAStaff Reading1.585m0.890m14. Find out the missing figures & complete the level book page. Apply using Arithmetic Check.B.SI.SF.SH.IR.LRemarks4.36XxPoint 1X192.000Point 23.9106.520XXPoint 35.390191.620B.M4.730XPoint 4X203.300Point 5 (StaffInverted)4.33xxXPoint 62.990194.830Point 7
15. The Following consecutive readings were taken with a level and a 4.0m staff on a continuouslysloping ground at a common interval of 30m: 0.78, 1.535, 1.955, 2.43, 2.985, 3.48, 1.155, 1.96,2.365, 3.64, 0.935, 1.045, 1.63, 2.545.The reduced level of First point A was 180.750m. Rule out the page of a level field book andenter the above readings. Calculate the reduced levels of the points by the collimation system(Height of Instrument Method) and the Rise & Fall method. Also calculate the gradient of theline joining the first and the last points.16. A gradient of 1 in 400 falling from elevation 67.45 was set out by driving pegs at 100m intervalswith its top of pegs on the required gradient. After a time it was suspected that some of thepegs had been disturbed and the following observations were taken for checking theirelevations. List the errors in the levels of the pegs if 7231.961.42Peg 140.9351.20Peg 261.50Peg 371.76Peg 482.03Peg 592.30Peg 6100.69Peg 72.59110.59Peg 8121.23Peg 9131.52Peg 1014150.611.211.7264.13B.M17. In running fly levels from a B.M of R.L 250m, the following readings (in m) were obtainedBacksight:1.315, 2.035, 1.980, 2.625Foresight:1.150, 3.450, 2.255From the last position
1. State the Objectives of Surveying? 2. What is basic principle on which Surveying has been classified? And explain them? 3. Differentiate between Plane Surveying & Geodetic Surveying? 4. State the various types of functional classification of Surveying? 5. State t
MYSTERY EXPRESSION GAME!!!!! i. Twice x Answer I: ii. 9 less than the Answer I Answer II: iii. The sum of Answer II and the product of 5 and t Answer III: iv. Answer III decreased by two-fifths the cube of another number Answer IV: v. Half of Answer IV Answer V: ANSWER V is E
Trigonometry Unit 4 Unit 4 WB Unit 4 Unit 4 5 Free Particle Interactions: Weight and Friction Unit 5 Unit 5 ZA-Chapter 3 pp. 39-57 pp. 103-106 WB Unit 5 Unit 5 6 Constant Force Particle: Acceleration Unit 6 Unit 6 and ZA-Chapter 3 pp. 57-72 WB Unit 6 Parts C&B 6 Constant Force Particle: Acceleration Unit 6 Unit 6 and WB Unit 6 Unit 6
1. Work the problem and find an answer. 2. Write your answer in the answer boxes at the top of the grid. Write your answer with the first digit in the left answer box OR with the last digit in the right answer box. Write only one digit or symbol in each answer box. Do NOT leave a blank answer box in the middle of an answer.
write each product on its corresponding answer line. Directions: Choose a division strategy to nd the quotient for each problem. Show your work and write each quotient on its corresponding answer line. 1. x Answer: 25 13 2. x 6 Answer: 1027 3. x 4 Answer: 827 4. Answer: 7) X 7 2 7 3 225 5 5. Answer: 6. Answer: 2457 7 116 8
applied results in task of automatic short answering. The method is: given a short answer (student answer or reference answer); the objective is to generate the paragraph vector of the answer. First, the answer is tokenized to get a list of its words and any necessary text processing is applied (e.g. removing punctuation marks).
AUD & NZD USD, 'carry', EUR The latest IMM data covers the week from 2 April to 09 April 2019 Stretched short Neutral Stretched long Abs. position Positioning trend EUR Short JPY Short GBP Short CHF Short CAD Short AUD Short NZD Short MXN Long BRL Short RUB Long USD* Long *Adjusted according to USD value of contracts
6.1 Developmental Plasticity: Answer Key 6.2 Depletion of California's Aquifers: Answer Key 6.3 Difference and Analytical Engines: Answer Key 6.4 The Black Death: Answer Key 6.5 South Asian Carnivores: Answer Key 6.6 Coal Mining: Answer Key 6.7 The Sixth Amendment: Answer Key 6.8 Signaling Theory: Answer Key
.34/Nonpolar region of phospholipid. Answer: C 35/Glycocalyx. Answer: A 36/Polar region of phospholipid. Answer: B 37/Peripheral protein. Answer: E 38/Integral protein. Answer: D 39/Identification "tags" for the cell. Answer: A 40/Receptors for signal transducers. Answer: D 41/Hydrophilic portion. Answer: B Match the following:
