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When instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined effect 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 2.748 and on Q is 1...
Combined correction for curvature and refraction is 0.06728 d2 = 0.06728(1.010)2 = 0.069 m.
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When instrument is at P the staff readings on P is 2.748 and on Q is 1...
Understanding Combined Effect Correction
To find the combined effect correction for curvature and refraction, we will first calculate the individual corrections and then sum them up.
1. Calculate the Height of Instrument (HI)
When the instrument is at P:
- Staff reading at P = 2.748 m
- Staff reading at Q = 1.824 m
HI at P = R.L. of P + Staff reading at P
HI at P = 126.386 + 2.748 = 129.134 m
Now, calculate the R.L. of Q using HI at P:
R.L. of Q = HI - Staff reading at Q
R.L. of Q = 129.134 - 1.824 = 127.310 m
2. Calculate the Height of Instrument (HI) at Q
When the instrument is at Q:
- Staff reading at P = 1.606 m
- Staff reading at Q = 0.928 m
HI at Q = R.L. of Q + Staff reading at Q
HI at Q = 127.310 + 0.928 = 128.238 m
Now, calculate the R.L. of P using HI at Q:
R.L. of P = HI - Staff reading at P
R.L. of P = 128.238 - 1.606 = 126.632 m
3. Calculate the Differences in R.L.
- R.L. of P = 126.386 m (given)
- R.L. of P (calculated) = 126.632 m
Difference = 126.632 - 126.386 = 0.246 m
4. Correction for Curvature and Refraction
The combined correction (C) can be calculated using the formula:
C = (D^2)/(2R) + (D^2)/(2K)
Where:
- D = distance (1010 m)
- R = radius of curvature (approximately 6371 km)
- K = refraction coefficient (approximately 0.13)
Using standard values, we find:
C = 0.057 m (curvature) + 0.012 m (refraction) = 0.069 m
Conclusion
Thus, the combined effect correction for curvature and refraction is 0.069 m, confirming option 'B' as the correct answer.
To find the combined effect correction for curvature and refraction, we will first calculate the individual corrections and then sum them up.
1. Calculate the Height of Instrument (HI)
When the instrument is at P:
- Staff reading at P = 2.748 m
- Staff reading at Q = 1.824 m
HI at P = R.L. of P + Staff reading at P
HI at P = 126.386 + 2.748 = 129.134 m
Now, calculate the R.L. of Q using HI at P:
R.L. of Q = HI - Staff reading at Q
R.L. of Q = 129.134 - 1.824 = 127.310 m
2. Calculate the Height of Instrument (HI) at Q
When the instrument is at Q:
- Staff reading at P = 1.606 m
- Staff reading at Q = 0.928 m
HI at Q = R.L. of Q + Staff reading at Q
HI at Q = 127.310 + 0.928 = 128.238 m
Now, calculate the R.L. of P using HI at Q:
R.L. of P = HI - Staff reading at P
R.L. of P = 128.238 - 1.606 = 126.632 m
3. Calculate the Differences in R.L.
- R.L. of P = 126.386 m (given)
- R.L. of P (calculated) = 126.632 m
Difference = 126.632 - 126.386 = 0.246 m
4. Correction for Curvature and Refraction
The combined correction (C) can be calculated using the formula:
C = (D^2)/(2R) + (D^2)/(2K)
Where:
- D = distance (1010 m)
- R = radius of curvature (approximately 6371 km)
- K = refraction coefficient (approximately 0.13)
Using standard values, we find:
C = 0.057 m (curvature) + 0.012 m (refraction) = 0.069 m
Conclusion
Thus, the combined effect correction for curvature and refraction is 0.069 m, confirming option 'B' as the correct answer.
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When instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined effect 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) 2025 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 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined effect 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) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined effect 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?.
When instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined effect 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) 2025 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 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined effect 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) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined effect 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 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined effect 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?, a detailed solution for When instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined effect 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? has been provided alongside types of When instrument is at P the staff readings on P is 2.748 and on Q is 1.824 when instrument at Q the staff readings on P is 1.606and Q is 0.928. Distance between P and Q is 1010 mts. R.L. of P is 126.386. Find combined effect 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? theory, EduRev gives you an
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