Civil Engineering - UPSC Civil Service Exam Questions - Discussion

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

[A]. 0.00527 m
[B]. 0.0527 m
[C]. 0.527 m
[D]. 0.673 m

Answer: Option B

Explanation:

No answer description available for this question.

Aryan said: (Feb 11, 2014)  
Error due to curvature and refraction = 0.0678*(d)^2.
Here, d=1 km.

Error= 0.0678.
Collimation error= 0.120 - 0.0678.

Kanu said: (Mar 7, 2017)  
Thanks @Aryan.

Anil said: (Mar 11, 2017)  
@Aryan.

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

Taofeek said: (Apr 17, 2017)  
@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.

Uttarayan said: (Jan 16, 2018)  
@Anil.

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

Mahendra Shukla said: (Dec 23, 2018)  
Given;

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

Akash said: (Sep 26, 2019)  
Collimation Error = 0.12-0.0673 = 0.0527.

Amit Singh said: (Jan 31, 2021)  
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).

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