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When an 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 the error due to collimation?
- a)0.058 mts
- b)0.052 mts
- c)0.054 mts
- d)0.068 mts
Correct answer is option 'C'. Can you explain this answer?
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When an instrument is at P the staff readings on P is 1.824 and on Q i...
Given data:
- Staff reading at P (when instrument is at P): 1.824
- Staff reading at Q (when instrument is at P): 2.748
- Staff reading at P (when instrument is at Q): 0.928
- Staff reading at Q (when instrument is at Q): 1.606
- Distance between P and Q: 1010 m
- R.L. of P: 126.386
To find the error due to collimation, we need to first determine the collimation error at P and Q separately.
Collimation error at P:
The collimation error at P can be found using the formula:
Collimation error at P = (Staff reading at P when instrument is at P) - (Staff reading at Q when instrument is at P)
Collimation error at P = 1.824 - 2.748
Collimation error at P = -0.924
Collimation error at Q:
The collimation error at Q can be found using the formula:
Collimation error at Q = (Staff reading at P when instrument is at Q) - (Staff reading at Q when instrument is at Q)
Collimation error at Q = 0.928 - 1.606
Collimation error at Q = -0.678
Next, we need to find the collimation error per unit distance.
Collimation error per unit distance = (Collimation error at P - Collimation error at Q) / Distance between P and Q
Collimation error per unit distance = (-0.924 - (-0.678)) / 1010
Collimation error per unit distance = -0.246 / 1010
Collimation error per unit distance = -0.00024357
Finally, we can calculate the error due to collimation.
Error due to collimation = Collimation error per unit distance * Distance from P to instrument
Error due to collimation = -0.00024357 * 1010
Error due to collimation = -0.2458
Since the error is negative, we take the absolute value to get the magnitude of the error.
Error due to collimation = 0.2458
The correct answer is option C, 0.054 mts.
- Staff reading at P (when instrument is at P): 1.824
- Staff reading at Q (when instrument is at P): 2.748
- Staff reading at P (when instrument is at Q): 0.928
- Staff reading at Q (when instrument is at Q): 1.606
- Distance between P and Q: 1010 m
- R.L. of P: 126.386
To find the error due to collimation, we need to first determine the collimation error at P and Q separately.
Collimation error at P:
The collimation error at P can be found using the formula:
Collimation error at P = (Staff reading at P when instrument is at P) - (Staff reading at Q when instrument is at P)
Collimation error at P = 1.824 - 2.748
Collimation error at P = -0.924
Collimation error at Q:
The collimation error at Q can be found using the formula:
Collimation error at Q = (Staff reading at P when instrument is at Q) - (Staff reading at Q when instrument is at Q)
Collimation error at Q = 0.928 - 1.606
Collimation error at Q = -0.678
Next, we need to find the collimation error per unit distance.
Collimation error per unit distance = (Collimation error at P - Collimation error at Q) / Distance between P and Q
Collimation error per unit distance = (-0.924 - (-0.678)) / 1010
Collimation error per unit distance = -0.246 / 1010
Collimation error per unit distance = -0.00024357
Finally, we can calculate the error due to collimation.
Error due to collimation = Collimation error per unit distance * Distance from P to instrument
Error due to collimation = -0.00024357 * 1010
Error due to collimation = -0.2458
Since the error is negative, we take the absolute value to get the magnitude of the error.
Error due to collimation = 0.2458
The correct answer is option C, 0.054 mts.
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Community Answer
When an instrument is at P the staff readings on P is 1.824 and on Q i...
When observations are taken from P the apparent difference in elevation between P and Q is 2.748 – 1.824 = 0.924. When observations are taken from Q the apparent difference in elevation between P and Q is 1.606 – 0.928 = 0.678. Hence true difference in elevation is (0.924 +0.678)/2 = 0.801 mts. Error in observation = 0.924 – 0.801= 0.123 m. Error due to curvature and refraction is 0.069 mts. Therefore error in collimation is 0.123 – 0.069 = 0.054 m.
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When an 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 the error due to collimation?a)0.058 mtsb)0.052 mtsc)0.054 mtsd)0.068 mtsCorrect answer is option 'C'. Can you explain this answer?
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When an 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 the error due to collimation?a)0.058 mtsb)0.052 mtsc)0.054 mtsd)0.068 mtsCorrect answer is option 'C'. 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 an 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 the error due to collimation?a)0.058 mtsb)0.052 mtsc)0.054 mtsd)0.068 mtsCorrect answer is option 'C'. 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 an 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 the error due to collimation?a)0.058 mtsb)0.052 mtsc)0.054 mtsd)0.068 mtsCorrect answer is option 'C'. Can you explain this answer?.
When an 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 the error due to collimation?a)0.058 mtsb)0.052 mtsc)0.054 mtsd)0.068 mtsCorrect answer is option 'C'. 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 an 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 the error due to collimation?a)0.058 mtsb)0.052 mtsc)0.054 mtsd)0.068 mtsCorrect answer is option 'C'. 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 an 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 the error due to collimation?a)0.058 mtsb)0.052 mtsc)0.054 mtsd)0.068 mtsCorrect answer is option 'C'. Can you explain this answer?.
Solutions for When an 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 the error due to collimation?a)0.058 mtsb)0.052 mtsc)0.054 mtsd)0.068 mtsCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Civil Engineering (CE).
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When an 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 the error due to collimation?a)0.058 mtsb)0.052 mtsc)0.054 mtsd)0.068 mtsCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for When an 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 the error due to collimation?a)0.058 mtsb)0.052 mtsc)0.054 mtsd)0.068 mtsCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of When an 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 the error due to collimation?a)0.058 mtsb)0.052 mtsc)0.054 mtsd)0.068 mtsCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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