Civil Engineering - UPSC Civil Service Exam Questions - Discussion

Discussion Forum : UPSC Civil Service Exam Questions - Section 2 (Q.No. 16)
16.
The distance between two bench marks is 1000 m. If during levelling, the total error due to collimation, curvature and refraction is found to be + 0.120 m, then the magnitude of the collimation error is
0.00527 m
0.0527 m
0.527 m
0.673 m
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
9 comments Page 1 of 1.

Anijith K S said:   2 years ago
C combined = 0.0673 * d² ( C combined = C refraction + C curvature)
C combined = 0.0673 * 1² = 0.0673.
C total = 0.12.
Then, C collimation = 0.12 - 0.0673 = 0.0527.
(4)

Amit Singh said:   5 years ago
Curvature correction(m) = -0.06749d^2 (d is in Km),
Refraction correction (m) = 0.01121d^2 (d is in Km),
Combined Correction (m) = -0.06728d^2 (d is in km).
(4)

Akash said:   6 years ago
Collimation Error = 0.12-0.0673 = 0.0527.
(3)

Mahendra shukla said:   7 years ago
Given;

Total error =collimation error + curvature error + refraction error.
.120=colli.error-.07857*1+.01122*1,
Colli.error = .120-.06735=.0527.
(1)

Uttarayan said:   8 years ago
@Anil.

The value of error due to curvature and refraction comes in m although d is taken in km.
(1)

Taofeek said:   8 years ago
@Anil.

The combined error due to curvature and refraction (ecomb) is thus given by ecomb = 0.0675 D2 m where D is the distance in km.

Anil said:   8 years ago
@Aryan.

How can you subtract? One is in km and other is in mt.

Kanu said:   8 years ago
Thanks @Aryan.

ARYAN said:   1 decade ago
Error due to curvature and refraction = 0.0678*(d)^2.
Here, d=1 km.

Error= 0.0678.
Collimation error= 0.120 - 0.0678.

Post your comments here:

Your comments will be displayed after verification.