Tiêu đề C2 MSTE Reviewer.pdf ✅

C2 MSTE Reviewer ^_^ 1. A surveyor measured a 700 m line using an uncalibrated tape that is known to be 0.015 m too long. He measured the line and came up with an error of -0.2099370189. Determine the tape length of the uncalibrated tape. A. 27 m C. 100 m B. 50 m D. 74 m Solution: Difficulty: 1/5 Correction, C = −emeasured distance C = −(−0.2099370189) = 0.2099370189 Correction due to uncalibrated tape length Too long: C = etape length × MD TL 0.2099370189 = 0.015 ( 700 TL ) TL = 50.015 m ≈ 50m 2. A group of Civil Engineering students was tasked to measure a 500 m walkway. The group has two (2) 50 m uncalibrated steel tapes, with one that is 0.013 m too short and the other that is 0.016 m too long. The group used the tapes alternately until they completed the task, starting with the 0.013 m too short. Determine the actual distance measured by the group. A. 499.985 m C. 498.053 m B. 501.641 m D. 502.164 m Solution: Difficulty: 2/5 Let: 0.013m too short Tape = T1 0.016m too long Tape = T2

C2 MSTE Reviewer ^_^ 4. A surveyor measured the area of a pentagonal lot using Double Meridian Distance Method (DMD). However, while calculating his data, he accidentally dropped his notes on wet soil. This caused smudges all over his data, leaving only some of his calculations legible as shown below: Line Latitude Departure DMD 2A AB 9.20 32.43 BC 70.11 -445.1985 CD -28.96 -26.8 DE 5.36 92.6744 EA -6.41 Determine the DMD of line EA A. -32.43 C. 133.0075 B. 0 D. 6.41 Solution: Difficulty: 1/5 DMD of last line = - (Departure of last line) DMDEA = −6.41 5. A 50 m tape weighing 1.075 kg has a standard pull of 8 kg. The tape’s cross-sectional area and modulus of elasticity are 0.05 cm2 and 200 GPa, respectively. What pull (normal tension) is required in order that the effect of sag will be eliminated when the tape is supported at the end points only? A. 162.42 N C. 149.12 N B. 152.84 N D. 197.40 N Solution: Difficulty: 1/5 Cpull = Csag (P2 − P1)L AE = w 2L 3 24P2 2
C2 MSTE Reviewer ^_^ (P2 − 8(9.81))(50000) (0.05 × 102)(200000) = ( 1.075(9.81) 50000 ) 2 (50000) 3 24P2 2 P2 = 197.4 N 6. The common tangent of a compound curve makes an angle of 12° from the tangent passing thru the P.C. and 18° from the tangent passing thru the P.T. If the radius of the second curve is 180 m, find the radius of the first curve if the length of the common tangent is 70 m long A. 354.804 m C. 324.352 m B. 394.76 m D. 263.501 m Solution: Difficulty: 1/5 T1 + T2 = CT R1 tan ( 12 2 ) + 180 tan ( 18 2 ) = 70 R1 = 394.76 m 7. Using Simpson’s 1/3 Rule, approximate the area of a piece of land having irregular boundaries as follows: Station Offset Distance (m) Station Offset Distance (m) 1+120 3.6 1+200 3.6 1+140 4.2 1+220 4.3 1+160 2.7 1+240 3.4 1+180 2.4 1+260 2.5 A. 480.33 m2 C. 444.67 m2 B. 460.67 m2 D. 473.00 m2 Solution: Difficulty: 1/5 A = 20 3 (3.6 + 4(4.2 + 2.4 + 4.3) + 2(2.7 + 3.6) + 3.4) + 3.4 + 2.5 2 = 480.33 m2