Difficulty: Medium
Correct Answer: 1.345
Explanation:
Introduction / Context:
This is a classic two-peg test scenario used to detect and correct the line of collimation error of a level. When the instrument is placed midway between pegs, the difference in staff readings gives the true difference in elevation between the pegs. When the instrument is placed near one peg, unequal sight lengths reveal the line-of-sight error, allowing computation of the corrected far staff reading.
Given Data / Assumptions:
Concept / Approach:
The line of collimation error produces a reading error proportional to sight length. Let e be the error per 100 m added to a reading (positive if it increases with distance). The difference in errors between B and A is e * (IB − IA) / 100. Matching observed and true differences determines e, which is then used to correct the far reading.
Step-by-Step Solution:
Verification / Alternative check:
Check using A's correction: near sight error = e * 0.1 = 0.005 m. Corrected difference = (1.39 − 0.045) − (1.11 − 0.005) = 1.345 − 1.105 = 0.240 m, matching Δ_true.
Why Other Options Are Wrong:
1.435 and 1.425 do not satisfy the true difference constraint when combined with the corrected near reading.
“None of these” is false because 1.345 is consistent with the two-peg correction.
Common Pitfalls:
Using distance difference 80 m incorrectly; forgetting to apply corrections to both near and far readings; sign confusion when deciding whether to add or subtract the error.
Final Answer:
1.345
Discussion & Comments