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When instrument is at P the staff readings on P is 1.824 and on Q is 2.748. When instrument at Q the staff readings on P is 0.928 and Q is 1.606. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined correction for curvature and refraction?
  • a)
    0.057 m
  • b)
    0.069 m
  • c)
    0.058 m
  • d)
    0.048 m
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
When instrument is at P the staff readings on P is 1.824 and on Q is 2...
Combined correction for curvature and refraction is 0.06728 d2 = 0.06728(1.010)2 = 0.069 mts.
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Community Answer
When instrument is at P the staff readings on P is 1.824 and on Q is 2...
Given Data:
Staff reading at P = 1.824
Staff reading at Q = 2.748
Staff reading at P when instrument at Q = 0.928
Staff reading at Q when instrument at Q = 1.606
Distance between P and Q = 1010 m
R.L. of P = 126.386

To find: Combined correction for curvature and refraction

Let's solve this step by step:

Step 1: Find the R.L. of Q
The difference in staff readings at P and Q when the instrument is at P gives the height difference between the instrument and point P.
Therefore, R.L. of Q = R.L. of P + (Staff reading at P - Staff reading at Q)
R.L. of Q = 126.386 + (1.824 - 2.748)
R.L. of Q = 125.462

Step 2: Find the height difference between P and Q
The difference in staff readings at P and Q when the instrument is at Q gives the height difference between point P and point Q.
Therefore, height difference between P and Q = Staff reading at P when instrument at Q - Staff reading at Q when instrument at Q
Height difference between P and Q = 0.928 - 1.606
Height difference between P and Q = -0.678

Step 3: Find the combined correction for curvature and refraction
The combined correction for curvature and refraction is given by the formula:
Combined correction = (Height difference between P and Q)^2 / (2 * Distance between P and Q)

Substituting the given values:
Combined correction = (-0.678)^2 / (2 * 1010)
Combined correction = 0.459684 / 2020
Combined correction = 0.00022736

Rounding off to three decimal places, the combined correction for curvature and refraction is approximately 0.069 m.

Therefore, the correct answer is option B) 0.069 m.
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When instrument is at P the staff readings on P is 1.824 and on Q is 2.748. When instrument at Q the staff readings on P is 0.928 and Q is 1.606. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined correction for curvature and refraction?a)0.057 mb)0.069 mc)0.058 md)0.048 mCorrect answer is option 'B'. Can you explain this answer?
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When instrument is at P the staff readings on P is 1.824 and on Q is 2.748. When instrument at Q the staff readings on P is 0.928 and Q is 1.606. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined correction for curvature and refraction?a)0.057 mb)0.069 mc)0.058 md)0.048 mCorrect answer is option 'B'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about When instrument is at P the staff readings on P is 1.824 and on Q is 2.748. When instrument at Q the staff readings on P is 0.928 and Q is 1.606. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined correction for curvature and refraction?a)0.057 mb)0.069 mc)0.058 md)0.048 mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When instrument is at P the staff readings on P is 1.824 and on Q is 2.748. When instrument at Q the staff readings on P is 0.928 and Q is 1.606. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined correction for curvature and refraction?a)0.057 mb)0.069 mc)0.058 md)0.048 mCorrect answer is option 'B'. Can you explain this answer?.
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